Find the relative extrema of
f(x) = -2x^2 + 4x + 3
I don't know how to use the First derivative test to find it.
take the first derivative and set it equal to zero.
-4x + 4 = 0
x = 1
sub that back into the function.
To use the First Derivative Test to find the relative extrema of a function, you'll need to follow these steps:
1. Take the derivative of the function f(x) to find its first derivative, denoted as f'(x).
In this case, the derivative of f(x) = -2x^2 + 4x + 3 can be found using the power rule and the constant rule:
f'(x) = 2(-2)x^1 + 4 = -4x + 4.
2. Set the first derivative, f'(x), equal to zero to find the critical points.
-4x + 4 = 0.
Solve for x: -4x = -4, then x = 1.
So, x = 1 is a critical point.
3. Now, you need to determine whether the function f(x) is increasing or decreasing around the critical point x = 1.
Substitute a value less than 1, like x = 0, into f'(x), and evaluate:
f'(0) = -4(0) + 4 = 4.
Since the result is positive, f(x) is increasing for x < 1.
Substitute a value greater than 1, like x = 2, into f'(x), and evaluate:
f'(2) = -4(2) + 4 = -4.
Since the result is negative, f(x) is decreasing for x > 1.
4. Based on the behavior of f(x) around the critical point x = 1, you can determine the relative extrema.
- If the function changes from increasing to decreasing at x = 1, then there is a local maximum at that point.
- If the function changes from decreasing to increasing at x = 1, then there is a local minimum at that point.
Since f(x) changes from decreasing to increasing at x = 1, the function has a local minimum at x = 1.
Now, you can find the y-coordinate of the local minimum by substituting x = 1 into the original function f(x):
f(1) = -2(1)^2 + 4(1) + 3 = -2 + 4 + 3 = 5.
Therefore, the relative extrema of f(x) = -2x^2 + 4x + 3 is a local minimum at (1, 5).