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?
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
8 answers
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
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
Well, it asked for the value of the function, not the coordinates of the point
12 for min
16 for max
12 for min
16 for max
oops. try
https://www.wolframalpha.com/input/?i=x%5E3+%E2%88%92+9x%5E2+%2B+24x+%E2%88%92+4
https://www.wolframalpha.com/input/?i=x%5E3+%E2%88%92+9x%5E2+%2B+24x+%E2%88%92+4
dang. Anonymous is right.
Always read the question carefully.
Always read the question carefully.
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.
= 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.
Yeah, it looks like the question was asking for the y value so, 12 and 16 were the correct answers.
Good, well the wording of the question fooled not only you but all three of us.