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# crucial values( f '' )?

what are the crucial values(f '') for

f(x)= 8x^(1/3) + x^(4/3)

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- whitesox09Lv 71 decade agoFavorite Answer
f'(x) = (1/3)*8*x^(-2/3) + (4/3)*x^(1/3)

f''(x) = (-2/9)*8*x^(-5/3) + (4/9)*x^(-2/3)

To find the critical values of f'', you need to set it equal to 0.

f''=0=(-2/9)*8*x^(-5/3) + (4/9)*x^(-2/3)

(2/9)*8*x^(-5/3) = (4/9)*x^(-2/3)

(16/9)x^(-5/3) = (4/9)x^(-2/3)

16x^(-5/3) = 4x^(-2/3)

4 = x^(-2/3 + 5/3)

4 = x

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