Question

Find the x-coordinates of any relative extrema and inflection point(s) for the function f of x equals 9 times x raised to the one third power plus 9 halves times x raised to the four thirds power . You must justify your answer using an analysis of f ′(x) and f ′′(x).

Answer 1

\[f(x) = 9x^\frac{ 1 }{ 3 } + \frac{ 9 }{ 2 }x^\frac{ 4 }{ 3 }\]

Answer 2

I have that \[f'(x) = 3x^\frac{ 1 }{ 3 }[x ^{-2} + 2]\]

Answer 3

@ganeshie8

Answer 4

So, I set f'(x) = 0 and I got \[x = 0, \pm \frac{ 1 }{ \sqrt{-2} }\]

Answer 5

http://www.wolframalpha.com/input/?i=%289x%5E%281%2F3%29+%2B+9%2F2*x%5E%284%2F3%29%29%27

Answer 6

I'm confused...

Answer 7

@jim_thompson5910 @amistre64

Answer 8

Hi, Jim_thomspon5910

Answer 9

\[\Large f(x) = 9x^{1/3} + \frac{9}{2}x^{4/3}\]

\[\Large f \ '(x) = 9(1/3)x^{1/3-1} + \frac{9}{2}*\frac{4}{3}x^{4/3-1}\]

\[\Large f \ '(x) = 3x^{-2/3} + 6x^{1/3}\]

\[\Large f \ '(x) = 3x^{1/3}\left(x^{-2} + 2\right)\]

so I'm getting the same f ' (x) as you did

Answer 10

Then set it to 0, yes?

Answer 11

if you set f ' (x) equal to zero, and solve for x, you do get

\[\Large x = 0 \text{ or } x = \pm \frac{1}{\sqrt{-2}}\]

so I agree there as well

Answer 12

The portion \(\Large \pm\frac{1}{\sqrt{-2}}\) aren't real numbers, so we can ignore these two solutions

Answer 13

Now either do a first derivative test or a second derivative test to see if a local min or max is at x = 0

Answer 14

The number line check?

Answer 15

yeah you can do that for a first derivative test

Answer 16

see if f ' (x) changes sign for a value less than 0, and for a value greater than 0

Answer 17

when it is less than 0, it is decreasing, when it is greater than 0, it is increasing

Answer 18

correct on both

Answer 19

So 0 is a min

Answer 20

when x < 0, f ' (x) < 0

when x > 0, f ' (x) > 0

we have a local min at x = 0

Answer 21

the min is at x = 0

the actual min value itself is unknown at this point

Answer 22

plug x = 0 back into f(x) to find the min value

Answer 23

I'm not thinking

if x = 0, then f ' (x) is undefined because of the x^(-2) portion. That leads to a division by zero error. So x = 0 cannot be a critical value.

The graph of f(x) also shows a vertical tangent at x = 0

Answer 24

Why plug 0 back into f(x)?

Answer 25

\[\Large f(x) = 9x^{1/3} + \frac{9}{2}x^{4/3}\]

\[\Large f \ '(x) = 9(1/3)x^{1/3-1} + \frac{9}{2}*\frac{4}{3}x^{4/3-1}\]

\[\Large f \ '(x) = 3x^{-2/3} + 6x^{1/3}\]

\[\Large f \ '(x) = 3x^{1/3}\left(x^{-\color{red}{1}} + 2\right)\]

Answer 26

oh my bad, I factored incorrectly