Question

Let \[ f(x) = x^3 + 15x^2 - 33x - 36 \] be a real-valued function. Which of the following statements is/are TRUE?

• A. $f(x)$ does not have a local maximum.
• B. $f(x)$ has a local maximum.
• C. $f(x)$ does not have a local minimum.
• D. $f(x)$ has a local minimum.

Hint:

For cubic polynomials, critical points are obtained from the quadratic derivative $f'(x)$. Use $f''(x)$ to classify them: negative $\Rightarrow$ maximum, positive $\Rightarrow$ minimum. A cubic can have both a local max and a local min.

Answer 1

The Correct Option is B

Step 1: Find critical points.
We first compute the derivative of $f(x)$: \[ f'(x) = \frac{d}{dx}(x^3 + 15x^2 - 33x - 36) = 3x^2 + 30x - 33. \] Factorize: \[ f'(x) = 3(x^2 + 10x - 11) = 3(x+11)(x-1). \] So, critical points are at \[ x = -11 \text{and} x = 1. \] Step 2: Use the second derivative test.
Compute $f''(x)$: \[ f''(x) = \frac{d}{dx}(3x^2 + 30x - 33) = 6x + 30. \] At $x = -11$: \[ f''(-11) = 6(-11) + 30 = -66 + 30 = -36<0, \] which indicates a local maximum. At $x = 1$: \[ f''(1) = 6(1) + 30 = 36>0, \] which indicates a local minimum. Step 3: Verify statements.
- (A) False, because $f(x)$ does have a local maximum at $x = -11$.
- (B) True, $f(x)$ has a local maximum at $x = -11$.
- (C) False, because $f(x)$ does have a local minimum at $x = 1$.
- (D) True, $f(x)$ has a local minimum at $x = 1$.
\[ \boxed{\text{Correct statements: (B) and (D).}} \]