Question

For:
f(x)=2x^2*lnx-13x^2

Find the critical point(s) of the function on its domain, then use the Second Derivative Test to determine
whether they correspond to local minima, local maxima, or neither.

I found the first derivative to be f'(x)= 4xln(x)-24x and a critical point at x=0, is there any other C.P's?

Answer 1

Okay i found the other c.p at x= \[x=e^6\] but how do i apply the second deriv test to determine max min?

Do i just find the second derivative, then plug in numbers to the left, right, and in between my c.p's?

Answer 2

\[f(x)=2x^2 \ln(x)-13x^2 \]\[f'(x)=(2x^2)'\ln(x)+2x^2(\ln(x))'-13(2)x^{2-1} \\ f'(x)=4x \ln(x)+2x^2 \frac{1}{x}-26x \\ f'(x)=4xln(x)+2x-26x \\ f'(x)=4x \ln(x)-24x\]
So your derivative looks right! :)
\[f'(x)=0 \text{ when } 4x \ln(x)-24x=0 \\ x \ln(x)-6 x=0 \text{ dividing 4 on both sides } x(\ln(x)-6)=0\]
So this is give you two numbers but only one of them is really a critical number.
x=0 isn't because it isn't in the domain of the original equation
\[\ln(x)=6 \\ x=e^6 \text{ is a critical number }\]

Answer 3

yes you need to find the second derivative

Answer 4

then plug in your critical number into f''

Answer 5

if you get a positive output then you have a min at x=e^(6)
if you get a negative output then you have a max at x=e^(6)

Answer 6

Is it because when f">0 it is concave up, indicating that there is a minimum value, and vice versa?

Answer 7

si
or is it
ci

I mean yes :p

Answer 8

as ugly as it will look
you won't need a calculator

Answer 9

have you found the second derivative?