Question

Please Calculus help?

Given the derivative, f ‘(x) = -6x2 + 12x

a.State the critical values where
f ‘(x) =0

b. Find the second derivative, f “(x)

c. Use the second derivative to test the critical values found in part a.

Analyze the results to determine whether these critical values produce a relative maximum or minimum.

I already have the second derivative. Just need a)

Answer 1

for a)
set f'(x) = 0 and solve for the x values to get your critical points.

Answer 2

b. pretty self explanatory.. just take the derivative of -6x^2 + 12x

Answer 3

c. input your critical values in the second derivative...

Answer 4

f ‘(x) = -6x^2 + 12x
given, f'(x) = 0
so
0 = -6x^2 + 12x
solve now =)

Answer 5

Another tip : If f''(x) is positive, x is a minimum. If f''(x) is negative, x is a maximum.

Answer 6

0=-6x(x-2)

Answer 7

is that right?

Answer 8

yeah

Answer 9

x = 2

Answer 10

is that the only point?

Answer 11

yes

Answer 12

input it into your second derivative to see if it is positive/negative.
positive = minimum
negative = maximum

Answer 13

great. thank you