Question

Find the x-coordinates of any relative extrema and inflection point(s) for the function f(x)=9x^1/3 +9/2 x^4/3 . You must justify your answer using an analysis of f '(x) and f "(x)

Answer 1

I'm solving it now bear with me its going to take a while @sbentleyaz

Answer 2

@3mar

Answer 3

@triciaal
If you could help here, I would be grateful!

Answer 4

@triciaal could you please help

Answer 5

\[\large\rm f(x)=9x^{1/3} +\frac92x^{4/3} \]What do you need help with?
Were you able to find the first derivative function?

Answer 6

@sbentleyaz
Please, pay attention here!

Answer 7

is this correct so far
we want to find f'(x) - the first derivative and f"(x)- the second derivative

f(x)= 9x^1/3+ 9/2x^4/3
f'(x)=3x^-2/3 + 6x^1/3
f''(x)=-2x^-5/3+2x^-2/3


f'(x) is used to find the local extrema and f''(x) is used to find the inflection points
set f'(x) equal to zero and solve. We get x=0 and x=-1/2. We can not have 0 though, because it is in the denominator and you can not divide by 0. therefore the x value of the relative extrema is x=-1/2.
substitute this into the original equation and you get
Relative Extrema: (-1/2, -5.3)
for the inflection points we set f"(x) equal to zero and solve
x=0 and x=-1
Again, x can not equal 0 due to it being in the denominator, so x=-1
we plug -1 into the original equation to find the y coordinate
so

Answer 8

this is what i have been working on

Answer 9

They only wanted the x-coordinate of your relative extrema.
So you did a bit more work than they asked for hehe :)

Answer 10

I think it's positive 1 for the inflection point.
Lemme check my work again.

Answer 11

Okay thank you

Answer 12

I think x=0 can be an inflection point actually...\[\large\rm 0=-2x^{-5/3}+2x^{-2/3}\]Factoring gives us,\[\large\rm 0=-2x^{-5/3}(1-x)\]
We have possible inflection points where the second derivative is `zero` or `undefined`.

We should do a line test to see which ones are actually inflection points.

Answer 13

so inflection at x=0 and x=-1

Answer 14

and for the relative extrema we do not have one at x=0 correct @zepdrix

Answer 15

Plugging in test points,

\(\large\rm f''(-1)=-2(-1)^{-5/3}(1--1)=positive\)
\(\large\rm f''(1/2)=-2\left(\frac12\right)^{-5/3}\left(1-\frac12\right)=negative\)
\(\large\rm f''(2)=-2(2)^{-5/3}(1-2)=positive\)

Answer 16

but the neative fraction exponent will create a denominator where 0 can not be subbed in

Answer 17

so how could 0 be an answer