Finding Absolute Extrema — Question 5

Question 5 - Solution

We are given: $$f(x) = x^3 - 3x + 1$$

(a) Find critical points:

$$f'(x) = 3x^2 - 3 = 3(x^2 - 1) = 3(x - 1)(x + 1)$$

Set derivative to zero: $$f'(x) = 0 \Rightarrow x = \pm 1$$

Critical points: $x = -1$, $x = 1$

(b) Evaluate function:

$$f(-2) = (-2)^3 - 3(-2) + 1 = -8 + 6 + 1 = -1$$ $$f(-1) = (-1)^3 - 3(-1) + 1 = -1 + 3 + 1 = 3$$ $$f(1) = 1 - 3 + 1 = -1$$ $$f(2) = 8 - 6 + 1 = 3$$

(c) Conclusion:

• Absolute maximum: $\boxed{f(-1) = f(2) = 3}$

• Absolute minimum: $\boxed{f(-2) = f(1) = -1}$

Graph of $f(x) = x^3 - 3x + 1$ on $[-2, 2]$: