25
Active contributors today

## How do you find the instantaneous rate of change for f(x)=(x^2-2)/(x-1) for x=2?

Nam D.
Featured 4 months ago

Please refer to the explanation below.

#### Explanation:

We have to take the derivative of the function $f \left(x\right)$, and then plug in $x = 2$ into $f ' \left(x\right)$.

$f \left(x\right) = \frac{{x}^{2} - 2}{x - 1}$

So, we have to use the quotient rule, which states that,

$\frac{d}{\mathrm{dx}} \left(\frac{u}{v}\right) = \frac{u ' v - u v '}{v} ^ 2$

Here, $u = {x}^{2} - 2$, $v = x - 1$

$\therefore f ' \left(x\right) = \frac{2 x \left(x - 1\right) - 1 \cdot \left({x}^{2} - 2\right)}{x - 1} ^ 2$

$f ' \left(x\right) = \frac{2 {x}^{2} - 2 x - {x}^{2} + 2}{x - 1} ^ 2$

$f ' \left(x\right) = \frac{{x}^{2} - 2 x + 2}{x - 1} ^ 2$

Plugging in $x = 2$, we get

$f ' \left(2\right) = \frac{{2}^{2} - 2 \cdot 2 + 2}{2 - 1} ^ 2$

$f ' \left(2\right) = \frac{4 - 4 + 2}{1} ^ 2$

$f ' \left(2\right) = \frac{2}{1}$

$f ' \left(2\right) = 2$

So, the instantaneous rate of change will be $2$.

## Find the coordinates of the points on the graph of y=3x^2-2x at which tangent line is parallel to the line y=10x?

Parabola
Featured 3 months ago

$\left(2 , 8\right)$

#### Explanation:

Two lines are parallel if they share the same slope.

Any line parallel to $y = 10 x$ will have the slope 10.

Now, we have: $y = 3 {x}^{2} - 2 x$

We try to find $\frac{\mathrm{dy}}{\mathrm{dx}}$

=>$\frac{d}{\mathrm{dx}} \left(y\right) = \frac{d}{\mathrm{dx}} \left(3 {x}^{2} - 2 x\right)$

We use the power rule:

$\frac{d}{\mathrm{dx}} \left({x}^{n}\right) = n {x}^{n - 1}$ where $n$ is a constant.

$\frac{\mathrm{dy}}{\mathrm{dx}} = 3 \cdot 2 {x}^{2 - 1} - 2 \cdot 1 {x}^{1 - 1}$

=>$\frac{\mathrm{dy}}{\mathrm{dx}} = 6 x - 2$

Now, we set $\frac{\mathrm{dy}}{\mathrm{dx}} = 10$

=>$10 = 6 x - 2$

=>$12 = 6 x$

=>$2 = x$

We plug this value in to get our $y$, or $f \left(x\right)$.

$3 {\left(2\right)}^{2} - 2 \left(2\right) = y$

=>$12 - 4 = y$

=>$8 = y$

Therefore, the point is at $\left(2 , 8\right)$

## How do you integrate int 1/sqrt(4x+8sqrtx-24)  using trigonometric substitution?

Eric S.
Featured 3 months ago

Use the substitution $\sqrt{x} + 1 = \sqrt{7} \sec \theta$.

#### Explanation:

Let

$I = \int \frac{1}{\sqrt{4 x + 8 \sqrt{x} - 24}} \mathrm{dx}$

Complete the square in the square root:

$I = \frac{1}{2} \int \frac{1}{\sqrt{{\left(\sqrt{x} + 1\right)}^{2} - 7}} \mathrm{dx}$

Apply the substitution $\sqrt{x} + 1 = \sqrt{7} u$:

$I = \int \frac{\sqrt{7} u - 1}{\sqrt{{u}^{2} - 1}} \mathrm{du}$

Rearrange:

$I = \sqrt{7} \int \frac{u}{\sqrt{{u}^{2} - 1}} \mathrm{du} - \int \frac{1}{\sqrt{{u}^{2} - 1}} \mathrm{du}$

Apply the substitution $u = \sec \theta$:

$I = \sqrt{7} \sqrt{{u}^{2} - 1} - \int \sec \theta d \theta$

Hence

$I = \sqrt{7 {u}^{2} - 7} - \ln | \sec \theta + \tan \theta | + C$

Reverse the last substitution:

$I = \sqrt{7 {u}^{2} - 7} - \ln | u + \sqrt{{u}^{2} - 1} | + C$

Reverse the first substitution:

$I = \sqrt{{\left(\sqrt{x} + 1\right)}^{2} - 7} - \ln | \left(\sqrt{x} + 1\right) + \sqrt{{\left(\sqrt{x} + 1\right)}^{2} - 7} | + C$

Simplify:

$I = \sqrt{{x}^{2} + 2 \sqrt{x} - 6} - \ln | \left(\sqrt{x} + 1\right) + \sqrt{{x}^{2} + 2 \sqrt{x} - 6} | + C$

## Evaluate the definite integral int_2^4(x-3)^3dx ?

Shwetank Mauria
Featured 3 months ago

Integral is $0$.

#### Explanation:

Let $x - 3 = t$ then $\mathrm{dx} = \mathrm{dt}$ and

${\int}_{2}^{4} {\left(x - 3\right)}^{3} \mathrm{dx} = {\int}_{-} {1}^{1} {t}^{3} \mathrm{dt}$

= ${\left[{t}^{4} / 4\right]}_{-} {1}^{1}$

= $\left[\frac{1}{4} - \frac{1}{4}\right] = 0$

Note that when $x = 2$, $t = - 1$ and when $x = 4$, $t = 1$.

Now observe the graph of ${\left(x - 3\right)}^{3}$, as given below

graph{(x-3)^3 [0.542, 5.542, -1.2, 1.3]}

Note that area of curve in the interval $\left[2 , 3\right]$ is symmetric w.r.t. area of curve in the interval $\left[3 , 4\right]$, but they are in opposite direction, Hence the integral is $0$.

## Using the first derivative, what are the critical points of the curve and the intervals where it's increasing and decreasing?

Somebody N.
Featured 2 months ago

See below.

#### Explanation:

The critical points are the points where the first derivative equals zero, or the point doesn’t exist.

This is a polynomial so it is continuous for all $x \in \mathbb{R}$

First derivative of $f \left(x\right) = {x}^{4} - 8 {x}^{3} + 18 {x}^{2} - 5$

$\frac{\mathrm{dy}}{\mathrm{dx}} \left(f \left(x\right)\right) = 4 {x}^{3} - 24 {x}^{2} + 36 x$

Equating this to zero:

$4 {x}^{3} - 24 {x}^{2} + 36 x = 0$

$4 x \left({x}^{2} - 6 x + 9\right) = 0$

$4 x = 0 \implies x = 0$

$\left({x}^{2} - 6 x + 9\right) = 0$

Factor:

${\left(x - 3\right)}^{2} = 0 \implies x = 3$

Putting these values in $f \left(x\right)$

$f \left(0\right) = - 5$

$f \left(3\right) = 22$

Critical Points:

$\textcolor{b l u e}{x = 0}$ and $\textcolor{b l u e}{x = 3}$

$\left(0 , - 5\right)$ and $\left(3 , 22\right)$

$f \left(x\right)$ is increasing where $f ' \left(x\right) > 0$

$f \left(x\right)$ is decreasing where $f ' \left(x\right) < 0$

$4 {x}^{3} - 24 {x}^{2} + 36 x > 0$

${x}^{3} - 6 {x}^{2} + 9 x > 0$

$x \left({x}^{2} - 6 x + 9\right) > 0$

$x {\left(x - 3\right)}^{2} > 0$

$0 < x < 3$ , $3 < x < \infty$

Increasing in $\left(0 , 3\right)$ and $\left(3 , \infty\right)$

$4 {x}^{3} - 24 {x}^{2} + 36 x < 0$

${x}^{3} - 6 {x}^{2} + 9 x < 0$

$x \left({x}^{2} - 6 x + 9\right) < 0$

$x {\left(x - 3\right)}^{2} < 0$

$- \infty < x < 0$

Decreasing in $\left(- \infty , 0\right)$

GRAPH:

## lim_(n->oo)(sum_(k=1)^n(sqrt(2n^2+k)/(2n^2+k)))?

Hammer
Featured 2 months ago

${\lim}_{n \to \infty} {\sum}_{k = 1}^{n} \frac{\sqrt{2 {n}^{2} + k}}{2 {n}^{2} + k} = \frac{1}{\sqrt{2}}$

#### Explanation:

We have

${\lim}_{n \to \infty} {\sum}_{k = 1}^{n} \frac{\sqrt{2 {n}^{2} + k}}{2 {n}^{2} + k} = {\lim}_{n \to \infty} {\sum}_{k = 1}^{n} \frac{1}{\sqrt{2 {n}^{2} + k}} =$

= lim_(n->oo) sum_(k=1)^n 1/n * 1/sqrt(2+k/n^2)

Now, let's try to remember the definition of an integral. This might seem strange, but it will come of great use.

The definite integral ${\int}_{a}^{b} f \left(x\right) \mathrm{dx}$ represents the $\textcolor{red}{\text{net area}}$ defined by the graph of the function $f \left(x\right)$ and the x-axis. One way to find the value of an integral is by aproximating it with infinitely many rectangles ("boxes") of equal or not width.

This concept is called a $\textcolor{red}{\text{Riemann Sum}}$.

There are infinitely many rectangles, therefore, if there are $n$ rectangles, we have to take the limit as $n \to \infty$.

Now, let's take the case where the width is constant between all boxes. If we define ${x}_{1} , {x}_{2} , \ldots , {x}_{n}$ to be some "marks" on the x-axis such that the width of the $i$-th rectangle is ${x}_{i + 1} - {x}_{i} = \Delta {x}_{i}$, then its lenght is $f \left({x}_{i}\right)$. Since the width is equal, $\Delta {x}_{i}$ is the same for all $i$'s. We can more easily call it color(red)(Delta.

The Riemann Sum aproximates an integral by being the sum of all the rectangles. In order to simplify this further, let's take the particular case where color(red)(a = 0 and color(red)(b=1. We have:

${\int}_{0}^{1} f \left(x\right) \mathrm{dx} = {\lim}_{n \to \infty} {\sum}_{i = 1}^{n} \Delta \cdot f \left({x}_{i}\right)$

This is only true if ${x}_{n}$ is equal to $1$.

One way to make $\Delta$ constant is to define ${x}_{k} = \frac{k}{n}$, for some $k$, $1 \le k \le n$. This way, ${x}_{n} = 1$ and

$\textcolor{red}{\Delta} = {x}_{i + 1} - {x}_{i} = \frac{i + 1}{n} - \frac{i}{n} = \textcolor{red}{\frac{1}{n}}$.

${\int}_{0}^{1} f \left(x\right) \mathrm{dx} = {\lim}_{n \to \infty} {\sum}_{i = 1}^{n} \frac{1}{n} \cdot f \left(\frac{i}{n}\right)$.

This looks familiar to our original sum, ${\lim}_{n \to \infty} {\sum}_{k = 1}^{n} \frac{1}{n} \cdot \frac{1}{\sqrt{2 + \frac{k}{n} ^ 2}}$.

All we are left to do is to assume they're equal and then solve the integral.

${\lim}_{n \to \infty} {\sum}_{k = 1}^{n} \frac{1}{n} \cdot \frac{1}{\sqrt{2 + \frac{k}{n} ^ 2}} = {\lim}_{n \to \infty} {\sum}_{i = 1}^{n} \frac{1}{n} \cdot f \left(\frac{i}{n}\right)$

$i$ and $k$ are just the indexes, the name of which doesn't matter. Both sums could be counting $i$ or $k$, it's irrelevant.

${\lim}_{n \to \infty} {\sum}_{k = 1}^{n} \frac{1}{n} \cdot \frac{1}{\sqrt{2 + \frac{k}{n} ^ 2}} = {\lim}_{n \to \infty} {\sum}_{k = 1}^{n} \frac{1}{n} \cdot f \left(\frac{k}{n}\right)$

$\implies f \left(\frac{k}{n}\right) = \frac{1}{\sqrt{2 + \frac{k}{n} ^ 2}} \implies f \left(x\right) = \frac{1}{\sqrt{2 + \frac{x}{n}}}$.

We have to find the integral of this function:

${\lim}_{n \to \infty} {\int}_{0}^{1} \frac{\mathrm{dx}}{\sqrt{2 + \frac{x}{n}}}$

Since $n \to \infty$, this means that $\frac{x}{n} = 0$, as $x$ is finite.

${\lim}_{n \to \infty} {\int}_{0}^{1} \frac{\mathrm{dx}}{\sqrt{2}} = \frac{1}{\sqrt{2}}$.

This is because it forms a rectangle with the x-axis, the width and lenght of which is $1$ and $\frac{1}{\sqrt{2}}$, respectively.

As a conclusion,

color(blue)(lim_(n->oo) sum_(k=1)^(n) sqrt(2n^2+k)/(2n^2+k) = 1/sqrt2.

##### Questions
• · An hour ago
• · An hour ago
• · An hour ago
• · 3 hours ago
• · 4 hours ago
• · 4 hours ago
• · 5 hours ago
• · 5 hours ago
• · 6 hours ago
• · 6 hours ago
• · 7 hours ago
• · 9 hours ago
• · 9 hours ago
• · 9 hours ago
• 10 hours ago · in Quotient Rule
• · 11 hours ago
• · 11 hours ago
• · 12 hours ago