# What are the critical points of h(x)=lnsqrt(3x-2(x^2))?

Feb 8, 2018

$\left(\frac{3}{4} , 0.059\right)$

#### Explanation:

Critical points exist when the derivative of the given point is 0 or undefined.

Let's find $h ' \left(x\right)$ first.

Remember the chain rule, power rule, and finding the derivative of $\ln x$

The chain rule states that:
If $f \left(x\right) = g \left(h \left(x\right)\right)$, then $f ' \left(x\right) = g ' \left(h \left(x\right)\right) \cdot h ' \left(x\right)$

The power rule states that $\frac{d}{\mathrm{dx}} \left({x}^{n}\right) = n {x}^{n - 1}$ when $n$ is a constant.

Also, $\frac{d}{\mathrm{dx}} \left(\ln x\right) = \frac{1}{x} \cdot \frac{d}{\mathrm{dx}} \left(x\right)$. Of course, $\frac{d}{\mathrm{dx}} \left(x\right)$ is just one.

Therefore,
$h ' \left(x\right) = \frac{1}{\sqrt{3 x - 2 {x}^{2}}} \cdot \frac{d}{\mathrm{dx}} \sqrt{3 x - 2 {x}^{2}}$

=>$h ' \left(x\right) = \frac{1}{\sqrt{3 x - 2 {x}^{2}}} \cdot \frac{1}{2} \cdot {\left(3 x - 2 {x}^{2}\right)}^{\frac{1}{2} - 1} \cdot \frac{d}{\mathrm{dx}} \left(3 x - 2 {x}^{2}\right)$

=>$h ' \left(x\right) = \frac{1}{\sqrt{3 x - 2 {x}^{2}}} \cdot \frac{1}{2} \cdot {\left(3 x - 2 {x}^{2}\right)}^{- \frac{1}{2}} \cdot \left(3 \cdot 1 {x}^{1 - 1} - 2 \cdot 2 {x}^{2 - 1}\right)$

=>$h ' \left(x\right) = \frac{1}{\sqrt{3 x - 2 {x}^{2}}} \cdot \frac{1}{2} \cdot {\left(3 x - 2 {x}^{2}\right)}^{- \frac{1}{2}} \cdot \left(3 - 4 x\right)$

=>$h ' \left(x\right) = \frac{1}{2 \left(\sqrt{3 x - 2 {x}^{2}}\right)} \cdot \frac{3 - 4 x}{\sqrt{3 x - 2 {x}^{2}}}$

=>$h ' \left(x\right) = \frac{3 - 4 x}{2 \left(3 x - 2 {x}^{2}\right)}$

=>$h ' \left(x\right) = \frac{3 - 4 x}{2 x \left(3 - 2 x\right)}$

Now, we want $h ' \left(x\right)$ to be 0 or undefined.

We have:
=>$0 = \frac{3 - 4 x}{2 x \left(3 - 2 x\right)}$

We see that $3 - 4 x = 0$

=>$0 = - 4 x + 3$

=>$- 3 = - 4 x$

=>$\frac{3}{4} = x$

For $\frac{3 - 4 x}{2 x \left(3 - 2 x\right)}$ to be undefined, $\left(2 x \left(3 - 2 x\right)\right)$ have to equal 0.

=>$\left(2 x \left(3 - 2 x\right)\right) = 0$ Either $2 x = 0$ or $3 - 2 x = 0$

When we solve these, we get: $0 = x = \frac{3}{2}$

Now, we check whether these $x$ values are valid for our original function.

When we plug this in, we see that $0$ and $\frac{3}{2}$ give us an undefined value.

However, when $x = \frac{3}{4}$, $h \left(x\right) \approx 0.059$

Therefore, the critical point is at $\left(\frac{3}{4} , 0.059\right)$

Feb 8, 2018

$\left(\frac{3}{4} , \ln \left(\frac{3 \sqrt{2}}{4}\right)\right)$
So critical point is at $x = \frac{3}{4}$

#### Explanation:

Set the derivative to zero to find critical points. To derive an ln function use this rule:

$\frac{d}{\mathrm{dx}} \ln \left(u\right) = \frac{u '}{u}$

To derive u to find u' use the general power rule:

$\frac{d}{\mathrm{dx}} {u}^{n} = n {\left(u\right)}^{n - 1} \cdot u$

So $\frac{d}{\mathrm{dx}} {\left(- 2 {x}^{2} + 3 x\right)}^{\frac{1}{2}}$=$\left(\frac{1}{2}\right) {\left(- 2 {x}^{2} + 3 x\right)}^{- \frac{1}{2}} \left(- 4 x + 3\right)$=$\frac{- 4 x + 3}{2 \sqrt{- 2 {x}^{2} + 3 x}}$=$u '$

Now we have u', so plug it into the derivative of ln equation:

$\frac{d}{\mathrm{dx}} \ln \left(- 2 {x}^{2} + 3 x\right) = \frac{\frac{- 4 x + 3}{2 \sqrt{- 2 {x}^{2} + 3 x}}}{\sqrt{- 2 {x}^{2} + 3}}$

Which after some algebra will give you :

$\frac{4 x - 3}{\left(2 x\right) \left(2 x - 3\right)}$

Find the zeroes of this to get the x value of the critical point(s) , which is only $x = \frac{3}{4}$ here. If you want asymptotes, set the denominator to zero and solve for x.