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
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
3 answers
yes I mean f(x)=9x^(1/3) + (9/2)x^(4/3). Sorry, I've been trying to figure this out for hours
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
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
1 answer