Question

Consider the following.
f(x) = x^3 − 9x^2 + 24x − 4
Find the local minimum and maximum value of f.
local minimum value
local maximum value
I tried to solve this and I got (4,12) for minimum and (2,16) for maximum but it was incorrect

Answer 1

if
y = = x^3 − 9x^2 + 24x − 4
y' = 3 x^2 -18 x + 24
y" = 6 x - 18

where does y' = 0, horizontal, max or min?
x^2 - 6 x + 8 = 0
(x-4)(x-2) = 0
x = 4 and 2
Now when x = 4
y" = 6*4-18 = 24-18 = + 6
so that is a minimum at x = 4
then y = 4^3 − 9* 4^2 + 24* 4 − 4 = 64-144+96-4 = 12
so min at (4,12)
agree with you
now at x = 2
y" = 6*2-18 = 12 - 18 = - 6 so yes, max there
y(2) = 2^3 − 9*2^2 + 24*2 − 4 = 8 -36 +48-4 = 16
Yes, max at (2,16)
Who says it is incorrect?

Answer 2

max/min when f'(x) = 0
f' = 3x^2 - 18x + 24 = 3(x-2)(x-4)
it's a min if f is concave up (f" > 0)
f" = 6x-18 = 6(x-3)
f"(2) < 0 so the max is at (2,16)
f"(4) > 0 so the min is at (4,12)

The answer key is wrong. The graph is very clear, at

https://www.wolframalpha.com/input/?i=x%5E3+%E2%88%92+9x%5E2+%2B+24x+%E2%88%92+4+for+x%3D2%2C4

Answer 3

Well, it asked for the value of the function, not the coordinates of the point
12 for min
16 for max

Answer 4

oops. try

https://www.wolframalpha.com/input/?i=x%5E3+%E2%88%92+9x%5E2+%2B+24x+%E2%88%92+4

Answer 5

dang. Anonymous is right.
Always read the question carefully.

Answer 6

f'(x) = 3x^2 - 18x + 24
= 0 for a max/min

3x^2 - 18x + 24 = 0
x^2 - 6x + 8 = 0
(x-2)(x-4) = 0
x = 2, x = 4

when x = 2, f(2) = 8 - 36 + 48 - 4 = 16
when x = 4, f(4) = 64 - 144 + 96 - 4 = 12

the 2nd derivative test would show that (2,16) is a max
and (4,16) is a min

looks like you were correct

check for a copy error of the original function,
or else the author of the question is wrong.

Answer 7

Yeah, it looks like the question was asking for the y value so, 12 and 16 were the correct answers.

Answer 8

Good, well the wording of the question fooled not only you but all three of us.