Question

Determine the value of the function
\[f(x) = 24x + 24x ^{3/2} - 16x ^{2}\] at the x-coordinate where the second derivative is zero.


Well, so far I've got the second derivative to be
\[f''(x) = 18x ^{-1/2}-32\] (which I am not even sure is correct)
but if it is, how would I find it where "the x-coordinate is zero"?

Answer 1

So you have:
\[f(x)=24x+24x^\frac{3}{2}+16x^2\]
So you have the derivative:
\[f'(x)=24+36x^\frac{1}{2}+32x\]
And the second derivative:
\[f''(x)=18x^{-\frac{1}{2}}+32=\frac{18}{\sqrt{x}}+32\]
So you want to find the x-value when \([f''(x)=0]\)

Answer 2

So then you would have:
\[\eqalign{
&f''(x)=0 \\
&\frac{18}{\sqrt{x}}+32=0 \\
&\frac{18}{\sqrt{x}}=-32 \\
&\frac{1}{\sqrt{x}}=-\frac{32}{18} \\
&\sqrt{x}=-\frac{18}{32} \\
&x=\left(-\frac{18}{32}\right)^2 \\
}\]

Answer 3

it's supposed to be a -32...

Answer 4

OOOOOOH OUCH. Sorry yeah
so then it would be:
\[\eqalign{
&f''(x)=0 \\
&\frac{18}{\sqrt{x}}-32=0 \\
&\frac{18}{\sqrt{x}}=32 \\
&\frac{1}{\sqrt{x}}=\frac{32}{18} \\
&\sqrt{x}=\frac{18}{32} \\
&x=\left(\frac{18}{32}\right)^2 \\
}\]

Answer 5

so x = 0.316?

Answer 6

okay, I got it, thank you! :)

Answer 7

Ehhh...yeah! About no prob :)