SHOW ALL WORK!!
the answer choices are
A. -1
B. 0
C.1
D. 4/3
E. 5/3
2007-12-31 14:02:05 · 7 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
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]
2007-12-31 14:09:13 · answer #1 · answered by ¿ /\/ 馬 ? 7 · 1⤊ 3⤋
its 0
2014-12-10 23:43:15 · answer #2 · answered by Md. 1 · 0⤊ 1⤋
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
2007-12-31 14:10:33 · answer #3 · answered by michael c 3 · 0⤊ 2⤋
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
2007-12-31 14:10:21 · answer #4 · answered by eyeshield42 3 · 0⤊ 1⤋
The answer is 0
2007-12-31 14:05:56 · answer #5 · answered by fatandsmooth 5 · 0⤊ 2⤋
English please
2007-12-31 14:05:18 · answer #6 · answered by ryan 6 · 0⤊ 3⤋
If this is for homework. Do it yourself. You will never learn anything if you don't try it yourself.
2007-12-31 14:04:39 · answer #7 · answered by Jeanny W 2 · 0⤊ 6⤋