Find the relative extrema and absolute extrema if any. Use the first derivative test for the function f(x)=x^3-3x+2
To find the relative extrema and absolute extrema of a function, we need to analyze its first and second derivatives. Let's follow these steps to find the extrema for the function f(x) = x^3 - 3x + 2.
Step 1: Find the first derivative of f(x) by differentiating it with respect to x:
f'(x) = 3x^2 - 3
Step 2: To find the critical points of f(x), set f'(x) equal to 0 and solve for x:
3x^2 - 3 = 0
Divide both sides by 3:
x^2 - 1 = 0
Factoring:
(x - 1)(x + 1) = 0
Solving for x, we obtain:
x = 1, x = -1
These critical points mark locations where a relative extremum could occur.
Step 3: Analyze the sign of the first derivative f'(x) between and outside of the critical points (x = -1, 1) to identify increasing and decreasing intervals.
To do this, we can pick test values within the intervals and evaluate f'(x) to determine its sign. We'll consider three intervals: (-∞, -1), (-1, 1), and (1, +∞).
For x < -1, we can choose x = -2:
f'(-2) = 3(-2)^2 - 3
= 12 - 3
= 9
Since f'(-2) is positive, the function is increasing in the interval (-∞, -1).
For -1 < x < 1, we can choose x = 0:
f'(0) = 3(0)^2 - 3
= -3
Since f'(0) is negative, the function is decreasing in the interval (-1, 1).
For x > 1, we can choose x = 2:
f'(2) = 3(2)^2 - 3
= 9
Since f'(2) is positive, the function is increasing in the interval (1, +∞).
Based on the first derivative test, we can conclude that there is a relative minimum at x = -1 and a relative maximum at x = 1.
Step 4: Determine the absolute extrema by considering the endpoints of the function's domain.
Since f(x) is a polynomial, it is defined for all real numbers (-∞, +∞). Therefore, we need to evaluate f(x) at the critical points and the endpoints of its domain.
f(-∞) = (-∞)^3 - 3(-∞) + 2
= -∞
f(-1) = (-1)^3 - 3(-1) + 2
= -2
f(1) = (1)^3 - 3(1) + 2
= 0
f(+∞) = (+∞)^3 - 3(+∞) + 2
= +∞
Analyzing these values, we see that f(-∞) and f(+∞) both tend towards infinity, but f(-2) reaches a relative minimum. Therefore, the absolute minimum of f(x) = x^3 - 3x + 2 is -2 at x = -1, and there is no absolute maximum.
To summarize, the function f(x) = x^3 - 3x + 2 has a relative minimum at x = -1 and a relative maximum at x = 1. The absolute minimum is -2 at x = -1, and there is no absolute maximum.