Set f'=0
x=0,-2,5 for f' to be zero.
maximum is when f" is negative
a) f"=(x+2)(x-5)+x(x-5)+(x(x+2)
at x=0, f"=-10 so x=0 is max
b. f" at x=-2
f"=(0+ -2(-7) = positive, so at x=-2, it is minimum
c. f" at x=5
f"=0+0+35 positive, local maximum
check my calcs
The derivative of a function is f'(x)=x(x+2)(x-5). Find the value of x at each point where f has a
(a) local maximum,
(b) local minimum, or
(c) point of inflection
2 answers
oops, when f" is 35, local min