Asked by Kait

Find the x-coordinates of any relative extrema and inflection point(s) for the function f(x)=9x^1/3+9/2x^4/3.

I've gotten -0.5 as the relative minimum and x=1 for the inflection points. Is this correct? If not, can you show me how to get the correct answer please

Answers

Answered by Reiny
Your lack of brackets leave the interpretation of the equation very ambiguous.

e.g. do you mean: f(x)=9x^(1/3) + (9/2)x^(4/3) ?

please retype before I attempt it
Answered by Kait
yes I mean f(x)=9x^(1/3) + (9/2)x^(4/3). Sorry, I've been trying to figure this out for hours
Answered by Reiny
Looks pretty straight forward ...

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

for relative max/min, f'(x) = 0
3x^(-2/3) + 6x^(1/3) = 0
3x^(-2/3) (1 + 2x) = 0
3x^(-2/3) = 0 ---> no solution
or
1+2x = 0
x = -1/2 or -.5 ------- you had that

for points of inflection,
-2x^(-5/3) + 2x^(-2/3) = 0
-2x^(-5/3) (1 + x) = 0
x = -1

I agree with both of your answers
There are no AI answers yet. The ability to request AI answers is coming soon!

Related Questions