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

Question

William Abdalla · Accepted Answer

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

Charles Arocha · Answer

#(3/4,0.059)#

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)#

Christopher Agresta · Answer

undefined To find the critical points of the function $ h(x) = \ln(\sqrt{3x - 2x^2}) $, we need to find where its derivative is equal to zero or undefined.

First, we find the derivative of $ h(x) $ using the chain rule and the derivative of the natural logarithm function:

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

Setting $ h'(x) $ equal to zero and solving for $ x $, we get:

$$
\frac{1}{\sqrt{3x - 2x^2}} \cdot \frac{1}{2} \cdot (3 - 4x) = 0
$$

This gives us two critical points: $ x = \frac{3}{4} $ and $ x = \frac{3}{2} $.

Now, we need to check for critical points where the derivative is undefined. The derivative is undefined when the denominator becomes zero. However, the expression $ \sqrt{3x - 2x^2} $ is always non-negative, so the derivative is never undefined.

Therefore, the critical points of $ h(x) $ are $ x = \frac{3}{4} $ and $ x = \frac{3}{2} $.