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

2 Answers
Feb 8, 2018

Answer:

#(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

Answer:

#(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.