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The derivative of f(x) = (x^4/3) – (x^5/5) attains its maximum value at x =?

SHOW ALL WORK!!the answer choices areA. -1B. 0C.1D. 4/3 E. 5/3

7 Answers

f(x) = (x^4/3) – (x^5/5)f'(x) = 12x^3/9 – 25x^4/25f'(x) = 4x^3/3 -x^4Solve max/min by setting f'(x)=0f'(x) = 4x^3/3 -x^40 = 4x^3/3 -x^40 = x^3 ((4/3)-x)x=0, 4/3Check if x=0 or x=4/3 is the max:f(x) = (x^4/3) – (x^5/5)f(0) = 0f(4/3) = ((4/3)^4/3) – ((4/3)^5/5) = ~.21Therefore, at x=0, there is a maximum.[Answer: B]... Show More
1st derivitive is (4/3)x^(1/3) – 1 it’s max is when that derivitive is = 02nd derivitive is (4/9)x^(-2/3)so x = 0 is when it is max... Show More
a) differentiate f(x) b) equate the dy/dx to 0, because when dy/dx=0 , f(x) has maximum value.c) solve the equation dy/dx = 0… to get the value of x... Show More
If this is for homework. Do it yourself. You will never learn anything if you don’t try it yourself.... Show More

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