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
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?
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
Well, it asked for the value of the function, not the coordinates of the point
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
dang. Anonymous is right.
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.
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.
To find the local minimum and maximum values of a function, we can use calculus. Here's how you can solve this problem step by step:
Step 1: Take the derivative of the function f(x) with respect to x.
f'(x) = 3x^2 - 18x + 24
Step 2: Set the derivative equal to zero to find critical points.
3x^2 - 18x + 24 = 0
Step 3: Solve the quadratic equation for x.
Using the quadratic formula, x = (-b ± sqrt(b^2 - 4ac)) / 2a, where a = 3, b = -18, and c = 24, we get:
x = (-(-18) ± sqrt((-18)^2 - 4(3)(24))) / 2(3)
Simplifying, we have:
x = (18 ± sqrt(324 - 288)) / 6
x = (18 ± sqrt(36)) / 6
x = (18 ± 6) / 6
x = 24/6 or 12/6
x = 4 or 2
Step 4: Use the second derivative test to determine whether each critical point is a local minimum or maximum.
To apply the second derivative test, we need to take the second derivative of f(x).
f''(x) = 6x - 18
For x = 4:
f''(4) = 6(4) - 18
f''(4) = 24 - 18
f''(4) = 6
Since the second derivative is positive (greater than zero), the function has a local minimum at x = 4.
For x = 2:
f''(2) = 6(2) - 18
f''(2) = 12 - 18
f''(2) = -6
Since the second derivative is negative (less than zero), the function has a local maximum at x = 2.
Step 5: Substitute the critical points back into the original function to find the corresponding y-values.
For the local minimum, when x = 4:
f(4) = (4)^3 - 9(4)^2 + 24(4) - 4
f(4) = 64 - 144 + 96 - 4
f(4) = 12
For the local maximum, when x = 2:
f(2) = (2)^3 - 9(2)^2 + 24(2) - 4
f(2) = 8 - 36 + 48 - 4
f(2) = 16
Therefore, the local minimum value is (4, 12) and the local maximum value is (2, 16).
Apologies for the incorrect answer you received earlier. I hope this explanation helps!