Question

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

Answer 1

`(3/4,0.059)`

Explanation:

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

Let's find `h'(x)` first.

Remember the chain rule, power rule, and finding the derivative of `lnx`

The chain rule states that:
If `f(x)=g(h(x))`, then `f'(x)=g'(h(x))*h'(x)`

The power rule states that `d/dx(x^n)=nx^(n-1)` when `n` is a constant.

Also, `d/dx(lnx)=1/x*d/dx(x)`. Of course, `d/dx(x)` is just one.

Therefore,
`h'(x)=1/(sqrt(3x-2x^2))*d/dxsqrt(3x-2x^2)`

=>`h'(x)=1/(sqrt(3x-2x^2))* 1/2*(3x-2x^2)^(1/2-1)*d/dx(3x-2x^2)`

=>`h'(x)=1/(sqrt(3x-2x^2))* 1/2*(3x-2x^2)^(-1/2)*(3*1x^(1-1)-2*2x^(2-1))`

=>`h'(x)=1/(sqrt(3x-2x^2))* 1/2*(3x-2x^2)^(-1/2)*(3-4x)`

=>`h'(x)=1/(2(sqrt(3x-2x^2)))*(3-4x)/sqrt(3x-2x^2)`

=>`h'(x)=(3-4x)/(2(3x-2x^2))`

=>`h'(x)=(3-4x)/(2x(3-2x))`

Now, we want `h'(x)` to be 0 or undefined.

We have:
=>`0=(3-4x)/(2x(3-2x))`

We see that `3-4x=0`

=>`0=-4x+3`

=>`-3=-4x`

=>`3/4=x`

For `(3-4x)/(2x(3-2x))` to be undefined, `(2x(3-2x))` have to equal 0.

=>`(2x(3-2x))=0` Either `2x=0` or `3-2x=0`

When we solve these, we get: `0=x=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 `3/2` give us an undefined value.

However, when `x=3/4`, `h(x)~~0.059`

Therefore, the critical point is at `(3/4,0.059)`

Answer 2

`(3/4,ln((3sqrt(2))/4))`
So critical point is at `x=3/4`

Explanation:

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

`d/dx ln(u) = (u')/u`

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

`d/dx u^n = n(u)^(n-1)*u`

So `d/dx(-2x^2+3x)^(1/2)`=`(1/2)(-2x^2+3x)^(-1/2)(-4x+3)`=`(-4x+3)/(2sqrt(-2x^2+3x))`=`u'`

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

`d/dx ln(-2x^2+3x) = ((-4x+3)/(2sqrt(-2x^2+3x)))/sqrt(-2x^2+3)`

Which after some algebra will give you :

`(4x-3)/((2x)(2x-3))`

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