find the absolute minimum and maximum valies of f(x)=-x^4+2x^3+5 on the interval [-2,2]
check f(-2) and f(2).
f' = -4x^3 + 6x^2
= -2x^2(2x-3)
so there are local extrema at x = 3/2
Not at x=0, since that's a double root of f'=0.
Now pick the max/min values of f at -2,3/2,2
To find the absolute minimum and maximum values of a function, we need to follow these steps:
1. Find the critical points of the function: These are the points where the derivative of the function is either zero or does not exist.
2. Evaluate the function at the critical points and at the endpoints of the interval to find the maximum and minimum values.
Let's start by finding the critical points:
1. Take the derivative of the function f(x) with respect to x:
f'(x) = -4x^3 + 6x^2
2. Set the derivative equal to zero and solve for x:
-4x^3 + 6x^2 = 0
Factor out x^2:
x^2(-4x + 6) = 0
Setting each factor equal to zero gives us two possible critical points:
x^2 = 0 => x = 0
-4x + 6 = 0 => x = 3/2
Next, we need to evaluate the function at the critical points and at the endpoints of the interval [-2, 2]:
f(-2) = -(-2)^4 + 2(-2)^3 + 5 = -16 + 16 + 5 = 5
f(2) = -(2)^4 + 2(2)^3 + 5 = -16 + 16 + 5 = 5
f(0) = -(0)^4 + 2(0)^3 + 5 = 0 + 0 + 5 = 5
f(3/2) = -(3/2)^4 + 2(3/2)^3 + 5 ≈ -5.0625 + 8.4375 + 5 ≈ 8.375
Finally, we compare the values we found to identify the absolute minimum and maximum:
- Absolute Minimum: The function has a minimum value of 0 at x = 0.
- Absolute Maximum: The function has a maximum value of approximately 8.375 at x = 3/2.
f ' (x) = - 4x^3 + 6x^2
= 0 for max/min
x^2(-4x + 6) = 0
x = 0 or x = 3/2
f(0) = 5
f(3/2) = -81/64 + 54/8 + 5 = 671/64 = appr 10.48
f(-2) = - 16 + 16 + 5 = 5
f(2) = -16 - 16 + 5 = -27
so the absolute max is 671/64 and the min is -27