Find absolute maximum and minimum for f(x)=(3x^3)−(18x^2)+27x−10 over the closed interval [2,4].
Thank you so much!!
f = 3x^3−18x^2+27x−10
f' = 9x^2 - 36x + 27 = 9(x^2-4x+3) = 9(x-1)(x-3)
so, f has a relative min/max at x=3
From what you know about the general shape of cubics, it should be clear that (3,-10) is a relative min.
since
f(2) = -4
f(4) = 2,
f(3) = -10 is the absolute min within the interval
f(4) = 2 is the absolute max in the interval
To find the absolute maximum and minimum of a function, we need to follow these steps:
1. Find the critical points: These are the points where the derivative of the function is either zero or undefined. To do this, we need to find the derivative of the given function f(x) = 3x^3 - 18x^2 + 27x - 10.
Taking the derivative, we get:
f'(x) = 9x^2 - 36x + 27
Now set f'(x) = 0 and solve for x:
9x^2 - 36x + 27 = 0
We can simplify this equation by dividing all terms by 9:
x^2 - 4x + 3 = 0
Factoring the equation, we get:
(x - 1)(x - 3) = 0
Solving for x, we find two critical points:
x₁ = 1
x₂ = 3
2. Evaluate the function at the critical points and endpoints: Now we need to calculate the function values at the critical points and the endpoints of the given interval [2, 4].
Evaluate f(x) at x = 2:
f(2) = 3(2^3) - 18(2^2) + 27(2) - 10
= 3(8) - 18(4) + 27(2) - 10
= 24 - 72 + 54 - 10
= -4
Evaluate f(x) at x = 4:
f(4) = 3(4^3) - 18(4^2) + 27(4) - 10
= 3(64) - 18(16) + 27(4) - 10
= 192 - 288 + 108 - 10
= 2
Evaluate f(x) at x = 1 (critical point):
f(1) = 3(1^3) - 18(1^2) + 27(1) - 10
= 3(1) - 18(1) + 27(1) - 10
= 3 - 18 + 27 - 10
= 2
Evaluate f(x) at x = 3 (critical point):
f(3) = 3(3^3) - 18(3^2) + 27(3) - 10
= 3(27) - 18(9) + 27(3) - 10
= 81 - 162 + 81 - 10
= -10
3. Determine the absolute maximum and minimum: The absolute maximum and minimum values of the function f(x) = 3x^3 - 18x^2 + 27x - 10 over the closed interval [2, 4] can be determined by comparing the function values at the critical points and endpoints.
The function values are:
f(2) = -4
f(4) = 2
f(1) = 2
f(3) = -10
We can see that the absolute maximum value is 2, and it occurs at x = 4 and x = 1.
The absolute minimum value is -10, and it occurs at x = 3.
Therefore, the absolute maximum of the function is 2 and the absolute minimum is -10 over the interval [2, 4].