GATE CS 2023 Q18 — Maxima and Minima, Mean Value Theorem, Integration

To find the local extrema of the function $f(x) = x^3 + 15x^2 - 33x - 36$, we first find its first derivative with respect to $x$:

$$f'(x) = 3x^2 + 30x - 33$$

Set $f'(x) = 0$ to find the critical points:

$$3(x^2 + 10x - 11) = 0$$

$$(x + 11)(x - 1) = 0$$

This gives the critical points at $x = -11$ and $x = 1$.

Next, we use the second derivative test to classify these critical points. The second derivative of $f(x)$ is:

$$f''(x) = 6x + 30$$

Evaluating $f''(x)$ at the critical points:

• At $x = -11$:
  $$f''(-11) = 6(-11) + 30 = -36 < 0$$
  Since the second derivative is negative, $f(x)$ has a local maximum at $x = -11$.
• At $x = 1$:
  $$f''(1) = 6(1) + 30 = 36 > 0$$
  Since the second derivative is positive, $f(x)$ has a local minimum at $x = 1$.

Thus, $f(x)$ has both a local maximum and a local minimum. Therefore, statements B and D are TRUE.