The derivative of f(x) = (x^4/3) – (x^5/5) attains its maximum value at x =?
Given the function ( f(x) = \frac{x^{4/3}}{1} - \frac{x^5}{5} ), at which value of ( x ) does the derivative attain its maximum value?
Please show all of your work!
The answer choices are:
A. -1
B. 0
C. 1
D. (\frac{4}{3})
E. (\frac{5}{3})
7 Answers
Dec 16, 2024
f(x) = (x^4/3) – (x^5/5)
f'(x) = 12x^3/9 – 25x^4/25
f'(x) = 4x^3/3 -x^4
Solve max/min by setting f'(x)=0
f'(x) = 4x^3/3 -x^4
0 = 4x^3/3 -x^4
0 = x^3 ((4/3)-x)
x=0, 4/3
Check if x=0 or x=4/3 is the max:
f(x) = (x^4/3) – (x^5/5)
f(0) = 0
f(4/3) = ((4/3)^4/3) – ((4/3)^5/5) = ~.21
Therefore, at x=0, there is a maximum.
[Answer: B]
1st derivitive is (4/3)x^(1/3) – 1
it’s max is when that derivitive is = 0
2nd derivitive is (4/9)x^(-2/3)
so x = 0 is when it is max
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
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