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
            
        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
    
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
    
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
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