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

2 Answers
Feb 8, 2018

(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)

Feb 8, 2018

(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.