# Make the internet a better place to learn

1

## How to solve lim_( xrarr1) 1/(x-1) - 3/(1-x^3)?

Alan P.
Featured 1 week ago

${\lim}_{x \rightarrow 1} \frac{1}{x - 1} - \frac{3}{1 - {x}^{3}}$ is $\textcolor{g r e e n}{\text{undefined}}$

#### Explanation:

Remember the cubic factoring:
$\textcolor{w h i t e}{\text{XXX}} \left({x}^{3} - 1\right) = \left(x - 1\right) \left({x}^{2} + x + 1\right)$

Replace
$\textcolor{w h i t e}{\text{XXX}} \frac{1}{x - 1}$
with the equivalent
$\textcolor{w h i t e}{\text{XXX}} \frac{1}{x - 1} \cdot \frac{\left(x - 1\right) \left({x}^{2} + x + 1\right)}{{x}^{3} - 1} = \frac{{x}^{2} + x + 1}{{x}^{3} - 1}$

and replace
$\textcolor{w h i t e}{\text{XXX}} - \frac{3}{1 - {x}^{3}}$
with the equivalent
$\textcolor{w h i t e}{\text{XXX}} - \frac{3}{1 - {x}^{3}} \cdot \frac{\left(- 1\right)}{\left(- 1\right)} = + \frac{3}{{x}^{3} - 1}$

$\frac{1}{x - 1} - \frac{3}{1 - {x}^{3}}$
becomes
$\textcolor{w h i t e}{\text{XXX}} \frac{\left({x}^{2} + x + 1\right) + 3}{{x}^{3} - 1} = \frac{{x}^{2} + x + 4}{{x}^{3} - 1}$

Note that as $x \rightarrow 1$ the denominator $\rightarrow 0$ (but the numerator does not, so L'Hopital's rule does not apply)

Since division by zero is undefined, this limit is undefined.

As an aid in verifying this result, here is a graph of $\frac{1}{x - 1} - \frac{3}{1 - {x}^{3}}$

Notice that whether this limit approaches $- \infty$ or $+ \infty$ depends upon which side of $1$ from which $x$ approaches.

1

## Find the derivative of y=√2x+3 from definition of derivative..?

Douglas K.
Featured 1 week ago

I guessed that you meant $y = \sqrt{2 x + 3}$.

#### Explanation:

The definition of a derivative is:

$y ' \left(x\right) = {\lim}_{h \to 0} \frac{y \left(x + h\right) - y \left(x\right)}{h}$

Given:

$y \left(x\right) = \sqrt{2 x + 3}$

Then:

$y \left(x + h\right) = \sqrt{2 \left(x + h\right) + 3}$

Let's find an equation for h.

Square both sides:

${\left(y \left(x + h\right)\right)}^{2} = 2 \left(x + h\right) + 3$

Use the distributive property:

${\left(y \left(x + h\right)\right)}^{2} = 2 x + 2 h + 3$

Please observe that $2 x + 3$ is the same as ${\left(y \left(x\right)\right)}^{2}$:

${\left(y \left(x + h\right)\right)}^{2} = {\left(y \left(x\right)\right)}^{2} + 2 h$

Subtract ${\left(y \left(x\right)\right)}^{2}$ from both side and flip the equation:

$2 h = {\left(y \left(x + h\right)\right)}^{2} - {\left(y \left(x\right)\right)}^{2}$

Divide both sides by 2:

$h = \frac{{\left(y \left(x + h\right)\right)}^{2} - {\left(y \left(x\right)\right)}^{2}}{2}$

Substitute this for h into the definition:

$y ' \left(x\right) = {\lim}_{h \to 0} \frac{y \left(x + h\right) - y \left(x\right)}{\frac{{\left(y \left(x + h\right)\right)}^{2} - {\left(y \left(x\right)\right)}^{2}}{2}}$

Flip the 2 up to the numerartor:

$y ' \left(x\right) = {\lim}_{h \to 0} \frac{2 \left(y \left(x + h\right) - y \left(x\right)\right)}{{\left(y \left(x + h\right)\right)}^{2} - {\left(y \left(x\right)\right)}^{2}}$

We know how the difference of two squares factors, $\left({a}^{2} - {b}^{2}\right) = \left(a + b\right) \left(a - b\right)$, so we can do this to the denominator:

$y ' \left(x\right) = {\lim}_{h \to 0} \frac{2 \left(y \left(x + h\right) - y \left(x\right)\right)}{\left(y \left(x + h\right) + y \left(x\right)\right) \left(y \left(x + h\right) - y \left(x\right)\right)}$

Please observe that a common factor cancels:

$y ' \left(x\right) = {\lim}_{h \to 0} \frac{2 \cancel{\left(y \left(x + h\right) - y \left(x\right)\right)}}{\left(y \left(x + h\right) + y \left(x\right)\right) \cancel{\left(y \left(x + h\right) - y \left(x\right)\right)}}$

Here is the equation with the cancelled factors removed:

$y ' \left(x\right) = {\lim}_{h \to 0} \frac{2}{y \left(x + h\right) + y \left(x\right)}$

Now it is ok to let $h \to 0$

$y ' \left(x\right) = \frac{2}{y \left(x\right) + y \left(x\right)}$

$y ' \left(x\right) = \frac{2}{2 y \left(x\right)}$

$y ' \left(x\right) = \frac{1}{y \left(x\right)}$

$y ' \left(x\right) = \frac{1}{\sqrt{2 x + 3}}$

4

## 3xtan^-1 5xdx=7tan5ydy?

Eddie
Featured 1 week ago

$\implies y = \frac{1}{5} {\cos}^{- 1} \left(\pm C {e}^{\left(- \frac{15}{14} {x}^{2} {\tan}^{-} 1 \left(5 x\right) + \frac{3}{14} x - \frac{3}{70} {\tan}^{- 1} \left(5 x\right)\right)}\right)$

#### Explanation:

Is it: $3 x {\tan}^{-} 1 \left(5 x\right) \setminus \mathrm{dx} = 7 \tan \left(5 y\right) \setminus \mathrm{dy}$ ?

If so this means that:

$3 x {\tan}^{-} 1 \left(5 x\right) \setminus \mathrm{dx} - 7 \tan \left(5 y\right) \setminus \mathrm{dy} = 0$

We can test if this is exact , ie if there exists (level surface) function $f \left(x , y\right) = C$ so that:

$\mathrm{df} = {f}_{x} \mathrm{dx} + {f}_{y} \mathrm{dy} \textcolor{red}{= 0}$

With ${f}_{x} = 3 x {\tan}^{-} 1 \left(5 x\right)$ and ${f}_{y} = - 7 \tan \left(5 y\right)$, we test the mixed partials:

${f}_{x y} = \frac{\partial}{\partial y} \left(3 x {\tan}^{-} 1 \left(5 x\right)\right) = 0$

${f}_{y x} = \frac{\partial}{\partial x} \left(- 7 \tan \left(5 y\right)\right) = 0$

${f}_{x y} = {f}_{y x}$ so the equation is exact and $f \left(x , y\right) = C$ exists.

Next we can look at:

${f}_{x} = 3 x {\tan}^{-} 1 \left(5 x\right)$

By IBP:

$\implies f \left(x , y\right) = \int 3 x {\tan}^{-} 1 \left(5 x\right) \setminus \mathrm{dx}$

$= \int {\left(\frac{3}{2} {x}^{2}\right)}^{p} r i m e {\tan}^{-} 1 \left(5 x\right) \setminus \mathrm{dx}$

$= \frac{3}{2} {x}^{2} {\tan}^{-} 1 \left(5 x\right) - \int \frac{3}{2} {x}^{2} {\left({\tan}^{-} 1 \left(5 x\right)\right)}^{p} r i m e \setminus \mathrm{dx}$

And as: $\frac{d}{\mathrm{dx}} \left({\tan}^{- 1} \left(\alpha x\right)\right) = \frac{\alpha}{{\alpha}^{2} {x}^{2} + 1}$

$= \frac{3}{2} {x}^{2} {\tan}^{-} 1 \left(5 x\right) - \int \frac{3}{2} {x}^{2} \textcolor{b l u e}{\frac{5}{25 {x}^{2} + 1}} \setminus \mathrm{dx}$

$= \frac{3}{2} {x}^{2} {\tan}^{-} 1 \left(5 x\right) - \frac{3}{10} \int \frac{25 {x}^{2} + 1 - 1}{25 {x}^{2} + 1} \setminus \mathrm{dx}$

$= \frac{3}{2} {x}^{2} {\tan}^{-} 1 \left(5 x\right) - \frac{3}{10} \int 1 - \textcolor{b l u e}{\frac{1}{25 {x}^{2} + 1}} \setminus \mathrm{dx}$

And as: $\int \frac{1}{{\alpha}^{2} {x}^{2} + 1} \setminus \mathrm{dx} = \frac{1}{\alpha} \int \frac{d}{\mathrm{dx}} \left({\tan}^{- 1} \left(\alpha x\right)\right) \setminus \mathrm{dx}$

$\implies f \left(x , y\right) = \frac{3}{2} {x}^{2} {\tan}^{-} 1 \left(5 x\right) - \frac{3}{10} x + \frac{3}{50} {\tan}^{- 1} \left(5 x\right) + \textcolor{red}{\eta \left(y\right)}$

Because we know that ${f}_{y} = - 7 \tan \left(5 y\right)$, we can test that against the partial of $f \left(x , y\right)$, ie:

${f}_{y} = \frac{\partial}{\partial y} \left(\frac{3}{2} {x}^{2} {\tan}^{-} 1 \left(5 x\right) - \frac{3}{10} x + \frac{3}{50} {\tan}^{- 1} \left(5 x\right) + \textcolor{red}{\eta \left(y\right)}\right) = \eta ' \left(y\right)$

$\implies \eta ' = - 7 \tan \left(5 y\right)$

So we integrate again:

$\eta \left(y\right) = - \int 7 \tan \left(5 y\right) \mathrm{dy}$

We know that: $\int \tan \left(\alpha y\right) \mathrm{dy} = - \frac{1}{\alpha} \ln \left\mid \cos \alpha y \right\mid + C$

$\implies \eta \left(y\right) = \frac{7}{5} \ln \left\mid \cos \left(5 y\right) \right\mid + C$, where $C$ is generic constant term.

This means that:

$f \left(x , y\right) = \frac{3}{2} {x}^{2} {\tan}^{-} 1 \left(5 x\right) - \frac{3}{10} x + \frac{3}{50} {\tan}^{- 1} \left(5 x\right) + \frac{7}{5} \ln \left\mid \cos \left(5 y\right) \right\mid = C$

OR:

$\frac{7}{5} \ln \left\mid \cos \left(5 y\right) \right\mid = - \frac{3}{2} {x}^{2} {\tan}^{-} 1 \left(5 x\right) + \frac{3}{10} x - \frac{3}{50} {\tan}^{- 1} \left(5 x\right) + C$

$\implies \left\mid \cos \right\mid \left(5 y\right) = {e}^{\left(- \frac{15}{14} {x}^{2} {\tan}^{-} 1 \left(5 x\right) + \frac{3}{14} x - \frac{3}{70} {\tan}^{- 1} \left(5 x\right) + C\right)}$

$= C {e}^{\left(- \frac{15}{14} {x}^{2} {\tan}^{-} 1 \left(5 x\right) + \frac{3}{14} x - \frac{3}{70} {\tan}^{- 1} \left(5 x\right)\right)}$

$\implies y = \frac{1}{5} {\cos}^{- 1} \left(\pm C {e}^{\left(- \frac{15}{14} {x}^{2} {\tan}^{-} 1 \left(5 x\right) + \frac{3}{14} x - \frac{3}{70} {\tan}^{- 1} \left(5 x\right)\right)}\right)$

2

## How do you prove that the limit of ((9-4x^2)/(3+2x))=6 as x approaches -1.5 using the epsilon delta proof?

Jim H
Featured 1 week ago

#### Explanation:

The explanation has two sections. There is a preliminary analysis to find the values used in the proof, then there is a presentation of the proof itself.

Finding the proof

By definition,

${\lim}_{x \rightarrow \textcolor{g r e e n}{a}} \textcolor{red}{f \left(x\right)} = \textcolor{b l u e}{L}$ if and only if

for every $\epsilon > 0$, there is a $\delta > 0$ such that:
for all $x$, $\text{ }$ if $0 < \left\mid x - \textcolor{g r e e n}{a} \right\mid < \delta$, then $\left\mid \textcolor{red}{f \left(x\right)} - \textcolor{b l u e}{L} \right\mid < \epsilon$.

We have been asked to show that

lim_(xrarrcolor(green)(-1.5))color(red)((9-4x^2)/(3+2x) = color(blue)(6)

So we want to make $\left\mid {\underbrace{\textcolor{red}{\frac{9 - 4 {x}^{2}}{3 + 2 x}}}}_{\textcolor{red}{f \left(x\right)}} - {\underbrace{\textcolor{b l u e}{6}}}_{\textcolor{b l u e}{L}} \right\mid$ less than some given $\epsilon$ and we control (through our control of $\delta$) the size of abs(x-underbrace(color(green)((-1.5))))_color(green)(a))

We want: abs((9-4x^2)/(3+2x) )-6) < epsilon

Look at the thing we want to make small. Rewrite this, looking for the thing we control.

$\left\mid \frac{9 - 4 {x}^{2}}{3 + 2 x} - 6 \right\mid = \left\mid \frac{\left(3 - 2 x\right) \left(3 + 2 x\right)}{3 + 2 x} - 6 \right\mid$

 = abs(((3-2x) - 6)

$= \left\mid - 2 x - 3 \right\mid$

Recall that we control the size of $\left\mid x - \left(- 1.5\right) \right\mid = \left\mid x - \left(- \frac{3}{2}\right) \right\mid$

I see abs(x+3 which is the same as $\left\mid x - \left(- 3\right) \right\mid$ so let's 'factor out $- 2$.

$= \left\mid - 2 \right\mid \left\mid x + \frac{3}{2} \right\mid$

$= 2 \left\mid x - \left(- \frac{3}{2}\right) \right\mid$

In order to make this less than $\epsilon$, it suffices to make $\left\mid x - \left(- 1.5\right) \right\mid$ less than $e \frac{\psi}{2}$

Proving our L is correct -- Writing the proof

Claim: ${\lim}_{x \rightarrow - 1.5} \left(\frac{9 - 4 {x}^{2}}{3 + 2 x}\right) = 6$

Proof:

Given $\epsilon > 0$, choose $\delta = \frac{\epsilon}{2}$. (Note that $\delta$ is positive.)

Now if < delta then

$\left\mid \frac{9 - 4 {x}^{2}}{3 + 2 x} - 6 \right\mid = \left\mid \frac{\left(3 - 2 x\right) \left(3 + 2 x\right)}{3 + 2 x} - 6 \right\mid$

 = abs(((3-2x) - 6)

$= \left\mid - 2 x - 3 \right\mid$

$= \left\mid - 2 \right\mid \left\mid x + \frac{3}{2} \right\mid$

$= 2 \left\mid x - \left(- \frac{3}{2}\right) \right\mid$

$< 2 \delta$

$= 2 \left(e \frac{\psi}{2}\right)$

$= \epsilon$

We have shown that for any positive $\epsilon$, there is a positive $\delta$ such that for all $x$, if $0 < \left\mid x - \left(- 1.5\right) \right\mid < \delta$, then $\left\mid \frac{9 - 4 {x}^{2}}{3 + 2 x} - 6 \right\mid < \epsilon$.

So, by the definition of limit, we have ${\lim}_{x \rightarrow - 1.5} \left(\frac{9 - 4 {x}^{2}}{3 + 2 x}\right) = 6$.

Note

This is an exampole of a limit in which the strict inequality $0 < \left\mid x - \delta \right\mid$ is very important. If we allowed $0 = \left\mid x - \left(- 1.5\right) \right\mid$, then a choice of $x = - 1.5$ would result in an undefined expression. $\left\mid \frac{9 - 4 {x}^{2}}{3 + 2 x} - 6 \right\mid$.

4

## How do you do? Solids of revolution

Steve
Featured 1 week ago

$4 \sqrt{3} \pi {a}^{3}$

#### Explanation:

We basically are being asked to calculate the volume of a spherical bead, that is, a sphere with a hole drilled through it.

Consider a cross section through the sphere, which we have centred on the origin of a Cartesian coordinate system:

The red circle has radius $2 a$, hence its equation is:

${x}^{2} + {y}^{2} = {\left(2 a\right)}^{2} \implies {x}^{2} + {y}^{2} = 4 {a}^{2}$

Method 1 - Calculate core and subtract from Sphere

First let us consider the volume of the entire Sphere, which has radius $2 a$. I will use the standard volume formula $V = \frac{4}{3} \pi {r}^{3}$

$\therefore {V}_{\text{sphere}} = \frac{4}{3} \pi {\left(2 a\right)}^{3}$
$\text{ } = \frac{4}{3} \pi 8 {a}^{3}$
$\text{ } = \frac{32}{3} \pi {a}^{3}$

For the bore we can consider a solid of revolution. (Note this is not a cylinder as it has a curved top and bottom from the sphere). We will use the method of cylindrical shells and rotate about $O y$, the shell formula is:

$V = 2 \pi \setminus {\int}_{\alpha}^{\beta} \setminus x f \left(x\right) \setminus \mathrm{dx}$

Also note that we require twice the volume because we have a portion above and below the $x$-axis. The shell volume of revolution about $O y$ for the bore is given by:

${V}_{\text{bore}} = 2 \pi \setminus {\int}_{0}^{a} \setminus x \left(2 \sqrt{4 {a}^{2} - {x}^{2}}\right) \setminus \mathrm{dx}$
$\text{ } = 2 \pi \setminus {\int}_{0}^{a} \setminus 2 x \setminus \sqrt{4 {a}^{2} - {x}^{2}} \setminus \mathrm{dx}$

We can evaluate using a substitution:

Let $u = 4 {a}^{2} - {x}^{2} \implies \frac{\mathrm{du}}{\mathrm{dx}} = - 2 x$
When $\left\{\begin{matrix}x = 0 \\ x = a\end{matrix}\right. \implies \left\{\begin{matrix}u = 4 {a}^{2} \\ u = 3 {a}^{2}\end{matrix}\right.$

And so:

${V}_{\text{bore}} = - 2 \pi \setminus {\int}_{4 {a}^{2}}^{3 {a}^{2}} \setminus \sqrt{u} \setminus \mathrm{du}$
$\text{ } = 2 \pi \setminus {\int}_{3 {a}^{2}}^{4 {a}^{2}} \setminus \sqrt{u} \setminus \mathrm{du}$
$\text{ } = 2 \pi \setminus {\left[{u}^{\frac{3}{2}} / \left(\frac{3}{2}\right)\right]}_{3 {a}^{2}}^{4 {a}^{2}}$

$\text{ } = \frac{4 \pi}{3} \setminus {\left[{u}^{\frac{3}{2}}\right]}_{3 {a}^{2}}^{4 {a}^{2}}$
$\text{ } = \frac{4 \pi}{3} \setminus \left({\left(4 {a}^{2}\right)}^{\frac{3}{2}} - {\left(3 {a}^{2}\right)}^{\frac{3}{2}}\right)$
$\text{ } = \frac{4}{3} \pi \setminus \left(8 {a}^{3} - 3 \sqrt{3} {a}^{3}\right)$
$\text{ } = \frac{32}{3} \pi {a}^{3} - 4 \sqrt{3} \pi {a}^{3}$

And so the total volume is given by:

 V_("total") = V_("sphere") - V_("bore" 
$\text{ } = \frac{32}{3} \pi {a}^{3} - \left(\frac{32}{3} \pi {a}^{3} - 4 \sqrt{3} \pi {a}^{3}\right)$
$\text{ } = 4 \sqrt{3} \pi {a}^{3}$

Method 2 - Calculate volume of bead directly

We can use the same method of a solid of revolution using the method of cylindrical shells and rotate about $O y$, but this time we will consider the bead itself rather than the core that is removed, again we need to twice the volume because we have a portion above and below the $x$-axis. The shell volume of revolution about $O y$ for the bead is given by:

${V}_{\text{total}} = 2 \pi \setminus {\int}_{a}^{2 a} \setminus x \left(2 \sqrt{4 {a}^{2} - {x}^{2}}\right) \setminus \mathrm{dx}$
$\text{ } = 2 \pi \setminus {\int}_{a}^{2 a} \setminus 2 x \setminus \sqrt{4 {a}^{2} - {x}^{2}} \setminus \mathrm{dx}$

We can evaluate using a substitution:

Let $u = 4 {a}^{2} - {x}^{2} \implies \frac{\mathrm{du}}{\mathrm{dx}} = - 2 x$
When $\left\{\begin{matrix}x = a \\ x = 2 a\end{matrix}\right. \implies \left\{\begin{matrix}u = 3 {a}^{2} \\ u = 0\end{matrix}\right.$

And so:

${V}_{\text{total}} = - 2 \pi \setminus {\int}_{3 {a}^{2}}^{0} \setminus \sqrt{u} \setminus \mathrm{du}$
$\text{ } = 2 \pi \setminus {\int}_{0}^{3 {a}^{2}} \setminus \sqrt{u} \setminus \mathrm{du}$
$\text{ } = 2 \pi \setminus {\left[{u}^{\frac{3}{2}} / \left(\frac{3}{2}\right)\right]}_{0}^{3 {a}^{2}}$

$\text{ } = \frac{4 \pi}{3} \setminus {\left[{u}^{\frac{3}{2}}\right]}_{0}^{3 {a}^{2}}$
$\text{ } = \frac{4 \pi}{3} \setminus \left({\left(3 {a}^{2}\right)}^{\frac{3}{2}} - 0\right)$
$\text{ } = \frac{4 \pi}{3} \setminus \left(3 \sqrt{3} {a}^{3}\right)$
$\text{ } = 4 \sqrt{3} \pi {a}^{3}$, as above

2

## (cosx+1/2).(1/3-cosx) what is the maximum value?

Cesareo R.
Featured 1 week ago

See below.

#### Explanation:

Calling $f \left(x\right) = \left(\frac{1}{2} + x\right) \left(\frac{1}{3} - x\right) = - {x}^{2} / 6 - \frac{x}{6} + \frac{1}{6}$ and

we know that the local minima/maxima are given at the stationary points or when

$\frac{d}{\mathrm{dx}} f \left(x\right) = 0$

but in our case we have

$f \left(g \left(x\right)\right)$ so the stationary points are when

$\frac{d}{\mathrm{dx}} f \left(g \left(x\right)\right) = - \frac{1}{6} \left(1 + 12 g \left(x\right)\right) g ' \left(x\right) = 0$

with $g \left(x\right) = \cos \left(x\right)$

so the stationary points are the solutions of

$\left\{\begin{matrix}1 + 12 g \left(x\right) = 1 + 12 \cos \left(x\right) = 0 \\ g ' \left(x\right) = - \sin \left(x\right) = 0\end{matrix}\right.$

and the stationary points are

${x}_{1} = - \text{arccos} \left(- \frac{1}{12}\right) + 2 k \pi$ and

${x}_{2} = k \pi$ for $k \in \mathbb{Z}$

The stationary points qualification is done using ${d}^{2} / \left({\mathrm{dx}}^{2}\right) f \left(g \left(x\right)\right)$

or

${d}^{2} / \left({\mathrm{dx}}^{2}\right) f \left(g \left(x\right)\right) = - 2 g ' {\left(x\right)}^{2} - \frac{1}{6} \left(1 + 12 g \left(x\right)\right) g ' ' \left(x\right)$

or

${d}^{2} / \left({\mathrm{dx}}^{2}\right) f \left(\cos \left(x\right)\right) = C o s \frac{x}{6} + 2 C o {s}^{2} \left(x\right) - 2 S {\in}^{2} \left(x\right)$

Regarding the stationary points the value of ${d}^{2} / \left({\mathrm{dx}}^{2}\right) f \left(\cos \left(x\right)\right)$qualifies the type of extremum. If ${d}^{2} / \left({\mathrm{dx}}^{2}\right) f \left(\cos \left(x\right)\right) > 0$ is a local minimum and if ${d}^{2} / \left({\mathrm{dx}}^{2}\right) f \left(\cos \left(x\right)\right) < 0$ is a local maximum.

so

${d}^{2} / \left({\mathrm{dx}}^{2}\right) f \left(\cos \left({x}_{1}\right)\right) = - \frac{143}{72}$ local maxima
${d}^{2} / \left({\mathrm{dx}}^{2}\right) f \left(\cos \left({x}_{2}\right)\right) = \frac{13}{6}$ local minima

Attached a plot showing this behavior.

so concluding, ${x}_{1}$ for the maxima and ${x}_{2}$ for the minima

NOTE. The same result for the maximum point would be obtained directly by making

$\frac{\mathrm{df}}{\mathrm{dx}} = - \frac{1}{6} - 2 x = 0 \to x = - \frac{1}{12}$ with the maximum value given by

$f \left(- \frac{1}{12}\right) = \frac{25}{144}$ and here

$\frac{{d}^{2} f}{{\mathrm{dx}}^{2}} \left(- \frac{1}{12}\right) = - 2$ qualifying a maximum.

3

## How do you integrate ln(x+sqrt(x^2+1)) ?

George C.
Featured 1 week ago

$\int \textcolor{w h i t e}{.} \ln \left(x + \sqrt{{x}^{2} + 1}\right) \textcolor{w h i t e}{.} \mathrm{dx} = x {\sinh}^{- 1} x - \sqrt{{x}^{2} + 1} + C$

$\textcolor{w h i t e}{\int \ln \left(x + \sqrt{{x}^{2} + 1}\right) \mathrm{dx}} = x \ln \left(x + \sqrt{{x}^{2} + 1}\right) - \sqrt{{x}^{2} + 1} + C$

#### Explanation:

Let us try a hyperbolic substitution.

Let $x = \sinh \theta$

Then:

$\int \textcolor{w h i t e}{.} \ln \left(x + \sqrt{{x}^{2} + 1}\right) \textcolor{w h i t e}{.} \mathrm{dx} = \int \textcolor{w h i t e}{.} \ln \left(\sinh \theta + \sqrt{{\sinh}^{2} \theta + 1}\right) \frac{\mathrm{dx}}{d \theta} d \theta$

$\textcolor{w h i t e}{\int \textcolor{w h i t e}{.} \ln \left(x + \sqrt{{x}^{2} + 1}\right) \textcolor{w h i t e}{.} \mathrm{dx}} = \int \textcolor{w h i t e}{.} \ln \left(\sinh \theta + \cosh \theta\right) \cosh \theta \textcolor{w h i t e}{.} d \theta$

$\textcolor{w h i t e}{\int \textcolor{w h i t e}{.} \ln \left(x + \sqrt{{x}^{2} + 1}\right) \textcolor{w h i t e}{.} \mathrm{dx}} = \int \textcolor{w h i t e}{.} \ln \left({e}^{\theta}\right) \cosh \theta \textcolor{w h i t e}{.} d \theta$

$\textcolor{w h i t e}{\int \textcolor{w h i t e}{.} \ln \left(x + \sqrt{{x}^{2} + 1}\right) \textcolor{w h i t e}{.} \mathrm{dx}} = \int \textcolor{w h i t e}{.} \theta \cosh \theta \textcolor{w h i t e}{.} d \theta$

$\textcolor{w h i t e}{\int \textcolor{w h i t e}{.} \ln \left(x + \sqrt{{x}^{2} + 1}\right) \textcolor{w h i t e}{.} \mathrm{dx}} = \theta \sinh \theta - \cosh \theta + C$

$\textcolor{w h i t e}{\int \textcolor{w h i t e}{.} \ln \left(x + \sqrt{{x}^{2} + 1}\right) \textcolor{w h i t e}{.} \mathrm{dx}} = x {\sinh}^{- 1} x - \sqrt{{x}^{2} + 1} + C$

If we would prefer not to have an inverse hyperbolic function in the answer, let's find another expression for it:

Let:

$y = {\sinh}^{- 1} x$

Then:

$x = \sinh y = \frac{1}{2} \left({e}^{y} - {e}^{- y}\right)$

Hence:

${e}^{y} - 2 x - {e}^{- y} = 0$

So:

${\left({e}^{y}\right)}^{2} - \left(2 x\right) \left({e}^{y}\right) - 1 = 0$

${e}^{y} = \frac{2 x \pm \sqrt{{\left(2 x\right)}^{2} + 4}}{2}$

$\textcolor{w h i t e}{{e}^{y}} = x \pm \sqrt{{x}^{2} + 1}$

If $y$ is real then ${e}^{y} > 0$ and we need the $+$ sign here.

So:

${e}^{y} = x + \sqrt{{x}^{2} + 1}$

and:

$y = \ln \left(x + \sqrt{{x}^{2} + 1}\right)$

That is:

${\sinh}^{- 1} x = \ln \left(x + \sqrt{{x}^{2} + 1}\right)$

So:

$\int \textcolor{w h i t e}{.} \ln \left(x + \sqrt{{x}^{2} + 1}\right) \textcolor{w h i t e}{.} \mathrm{dx} = x \ln \left(x + \sqrt{{x}^{2} + 1}\right) - \sqrt{{x}^{2} + 1} + C$

1

## How do you sketch the graph f(x)=2x^3-12x^2+18x-1?

HSBC244
Featured yesterday

Step $1$: Determine the first derivative

$f ' \left(x\right) = 6 {x}^{2} - 24 x + 18$

Step $2$: Determine the critical numbers

These will occur when the derivative equals $0$.

$0 = 6 {x}^{2} - 24 x + 18$

$0 = 6 \left({x}^{2} - 4 x + 3\right)$

$0 = \left(x - 3\right) \left(x - 1\right)$

$x = 3 \mathmr{and} 1$

Step $3$: Determine the intervals of increase/decrease

We select test points.

Test point 1: $x = 0$

$f ' \left(0\right) = 6 {\left(0\right)}^{2} - 24 \left(0\right) + 18 = 18$

Since this is positive, the function is uniformly increasing on $\left(- \infty , 1\right)$.

Test point 2: $x = 2$

$f ' \left(2\right) = 6 {\left(2\right)}^{2} - 24 \left(2\right) + 18 = - 6$

Since this is negative, the function is decreasing on $\left(1 , 3\right)$.

I won't select a test point for $\left(3 , \infty\right)$ because I know the function is increasing on the interval. The point $x = 3$ is referred to as a turning point because it goes from decreasing to increasing or vice versa.

Step $4$: Determine the second derivative

This is the derivative of the first derivative.

$f ' ' \left(x\right) = 12 x - 24$

Step $5$: Determine the points of inflection

These will occur when $f ' ' \left(x\right) = 0$.

$0 = 12 x - 24$

$0 = 12 \left(x - 2\right)$

$x = 2$

Step $6$: Determine the intervals of concavity

Once again, we select test points.

Test point $1$: $x = 1$

$f ' ' \left(1\right) = 12 \left(1\right) - 24 = - 12$

This means that $f \left(x\right)$ concave down (Since it's negative) on$\left(- \infty , 2\right)$.

This also means that $f \left(x\right)$ is concave up on $\left(2 , \infty\right)$.

Step $7$: Determine the x/y- intercept

$f \left(x\right)$ doesn't have any rational factors, so we'll forget about the x-intercepts (as a result, you would need Newton's Method or a similar method to find the x-intercepts).

The y-intercept is

$f \left(0\right) = 2 {\left(0\right)}^{3} - 12 {\left(0\right)}^{2} + 18 x - 1 = - 1$

If you can't connect the graph, you could always make a table of values. In the end, you should get a graph similar to the following. graph{2x^3- 12x^2 + 18x - 1 [-10, 10, -5, 5]}

Hopefully this helps!

2

## If u_n = int (sin nx)/sinx dx, >= 2, prove that u_n = (2sin(n-1)x)/(n-1)+u_(n-2) Hence evaluate: int_0^(pi/2) (sin5x)/ sinx dx?

Ratnaker Mehta
Featured 1 week ago

For the ${2}^{n d}$ Proof, refer to the Explanation.

#### Explanation:

Here is a Second Method to prove the Result for

$n \ge 2 , n \in \mathbb{N} .$

${u}_{n} = \int \sin \frac{n x}{\sin} x \mathrm{dx} .$

Now, $\sin \frac{n x}{\sin} x = \frac{1}{\sin} x \left\{\sin \left(\left(n x - 2 x\right) + 2 x\right)\right\}$

Knowing that, $\sin \left(A + B\right) = \sin A \cos B + \cos A \sin B ,$ we get,

$\sin \frac{n x}{\sin} x = \frac{1}{\sin} x \left\{\sin \left(n x - 2 x\right) \cos 2 x + \cos \left(n x - 2 x\right) \sin 2 x\right\}$

$= \frac{1}{\sin} x \left\{\left(\sin \left(\left(n - 2\right) x\right)\right) \left(1 - 2 {\sin}^{2} x\right) + \left(\cos \left(\left(n - 2\right) x\right)\right) \left(2 \sin x \cos x\right)\right\}$

$= \frac{1}{\sin} x \left\{\sin \left(\left(n - 2\right) x\right) - 2 {\sin}^{2} x \sin \left(\left(n - 2\right) x\right) + 2 \sin x \cos x \cos \left(\left(n - 2\right) x\right)\right\}$

$= \sin \frac{\left(n - 2\right) x}{\sin} x - 2 \sin x \sin \left(\left(n - 2\right) x\right) + 2 \cos x \cos \left(\left(n - 2\right) x\right)$

$= \sin \frac{\left(n - 2\right) x}{\sin} x + 2 \left\{\cos x \cos \left(\left(n - 2\right) x\right) - \sin x \sin \left(\left(n - 2\right) x\right)\right\}$

$= \sin \frac{\left(n - 2\right) x}{\sin} x + 2 \left\{\cos \left(\left(n - 2\right) x + x\right)\right\}$

$\therefore \sin \frac{n x}{\sin} x = \sin \frac{\left(n - 2\right) x}{\sin} x + 2 \cos \left(\left(n - 1\right) x\right) .$

$\Rightarrow {u}_{n} = \int \sin \frac{n x}{\sin} x \mathrm{dx} = \int \left\{\sin \frac{\left(n - 2\right) x}{\sin} x + 2 \cos \left(\left(n - 1\right) x\right)\right\} \mathrm{dx}$

$= \int \sin \frac{\left(n - 2\right) x}{\sin} x \mathrm{dx} + 2 \int \cos \left(\left(n - 1\right) x\right) \mathrm{dx} .$

"Hence, "u_n=u_(n-2)+(2sin((n-1)x))/(n-1)+C; n in NN, n>=2.#

Rest of the Soln. is the same as in the First Method.

Enjoy Maths.!

1

## A closed rectangular box is to have a width equal to the length and a surface area of 600 in3. What is the height of the box that maximizes the volume?

HSBC244
Featured 6 days ago

A height of $10$ inches would maximize the volume. See below for details.

#### Explanation:

Let $l$ be the length, $w$ be the width and $h$ be the height. However, since $l = w$, we can narrow ourselves down to two variables, let them be $l$ and $h$.

The formula for surface area of a rectangular prism is generally:

$S . A = 2 l h + 2 l w + 2 h w$

In this case, it becomes

$600 = 2 l h + 2 {l}^{2} + 2 l h$

$600 = 4 l h + 2 {l}^{2}$

We want to solve for one of the variables now.

$600 - 2 {l}^{2} = 4 l h$

$\frac{600 - 2 {l}^{2}}{4 l} = h$

$\frac{300 - {l}^{2}}{2 l} = h$

The volume of a rectangular prism is given by:

$V = l \times w \times h$

Which in our case becomes:

$V = {l}^{2} \times h$

Now:

$V = {l}^{2} \frac{300 - {l}^{2}}{2 l}$

$V = \frac{1}{2} l \left(300 - {l}^{2}\right)$

$V = \frac{1}{2} \left(300 l - {l}^{3}\right)$

$V = 150 l - \frac{1}{2} {l}^{3}$

Now find the derivative of this with respect to volume.

$V ' = 150 - \frac{3}{2} {l}^{2}$

Find the critical values. These will occur when the derivative equals $0$.

$0 = 150 - \frac{3}{2} {l}^{2}$

$\frac{3}{2} {l}^{2} = 150$

${l}^{2} = 100$

$l = \pm 10$

A negative side length doesn't make sense, so we don't accept it as a solution. We now verify graphically to insure this is a maximum.

And it is!

We realize the y-coordinate of the maximum is $y = 1000$, which is the maximum volume possible. All that is left for us to do is find the height that produces the maximum volume.

$V = {l}^{2} \cdot h$

$1000 = {\left(10\right)}^{2} \cdot h$

$1000 = 100 h$

$h = 10$

Hopefully this helps!