Consider the function f(x)=5–4x^2 , -3≤x≤1
The absolute maximum value is __________
and this occurs at x equals ___________
The absolute minimum value is _________
and this occurs at x equals ___________
To find the absolute maximum and minimum values of a function, we need to evaluate the function at critical points and endpoints within the given interval.
1. Critical Points:
To find critical points, we need to find where the derivative of the function is equal to zero or undefined. Let's start by finding the derivative of the function f(x):
f(x) = 5 - 4x^2
f'(x) = (d/dx)(5) - (d/dx)(4x^2)
f'(x) = 0 - 8x
f'(x) = -8x
To find critical points, set the derivative equal to zero:
-8x = 0
x = 0
Since the interval -3 ≤ x ≤ 1, x = 0 falls within this interval and is a critical point that we need to evaluate.
2. Endpoints:
Now let's evaluate the function at the endpoints of the interval, -3 and 1.
When x = -3:
f(-3) = 5 - 4(-3)^2
f(-3) = 5 - 4(9)
f(-3) = 5 - 36
f(-3) = -31
When x = 1:
f(1) = 5 - 4(1)^2
f(1) = 5 - 4(1)
f(1) = 5 - 4
f(1) = 1
Now we compare the values obtained at the critical point and endpoints to determine the absolute maximum and minimum.
Absolute Maximum:
The absolute maximum value is the largest value among the critical point and endpoints. In this case, the largest value is 5.
Therefore, the absolute maximum value is 5, and it occurs at x = 1.
Absolute Minimum:
The absolute minimum value is the smallest value among the critical point and endpoints. In this case, the smallest value is -31.
Therefore, the absolute minimum value is -31, and it occurs at x = -3.