Question

please help! find the absolute max and min of f(x)=x^3-12x on [0,4]

Answer 1

answer : 2

Answer 2

y(dx/dy) = 3x^2 -12
getting the critical numbers:
3x^2 - 12 = 0
3x^2 = 12
x^2 = 12/3
x^2 = 4
x = +-2

substitute it to the original value
y1=(2)^3-12(2)
y1=-16
y2=(-2)^3-12(-2)
y2=16
get the y value of the interval
y3 = (0)^3 - 12(0)
y3 =0
y4 = (4)^3 - 12(4)
y4 = 16

biggest is 16 now we know that abs max is 16 at x = 4 and x = -2
smalles is -16 so abs min is -16 at x=2

Answer 3

find the derivative of f(x) and solve it when it's equal to 0
f'(x)=3x^2-12
\[3x^2-12=0 \implies 3x^2=12 \implies x^2=4\]
x=2 or x=-2 but -2 is not in [0,4]
all you have to do now is to check the values of the function at x=2, and at the border points x=0, x=4
the maximum among them is the maximum in the interval, and same is applied to the minimum

Answer 4

oh yea i forgot x= -2 cant be considered~~ but nevertheless 16 is still the abs maximum ^^

Answer 5

f(2)=2^3-12(2)=8-24=-16
f(0)=0
f(4)=4^3-12(4)=16
so the absolute maximum is 16 at x=4 and the absolute minimum is -16 at x=2