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

To determine whether the given function \( f(x) = x^3 + 15x^2 - 33x - 36 \) has local extrema (local maximum or minimum), we need to find the critical points by setting the derivative equal to zero and then use the second derivative test.

1. First, find the first derivative of \( f(x) \):

\(f'(x) = \frac{d}{dx}(x^3 + 15x^2 - 33x - 36) = 3x^2 + 30x - 33\) 

1. Find the critical points by setting the first derivative equal to zero:

\(3x^2 + 30x - 33 = 0\)

This is a quadratic equation in the form of \(ax^2 + bx + c = 0\) with \(a = 3\), \(b = 30\), and \(c = -33\).

Using the quadratic formula, \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), to find the roots:

\(x = \frac{-30 \pm \sqrt{30^2 - 4 \cdot 3 \cdot (-33)}}{2 \cdot 3}\)

\(x = \frac{-30 \pm \sqrt{900 + 396}}{6}\)

\(x = \frac{-30 \pm \sqrt{1296}}{6}\)

\(x = \frac{-30 \pm 36}{6}\)

This gives us two solutions:

\(x_1 = \frac{6}{6} = 1\)

\(x_2 = \frac{-66}{6} = -11\)

1. Use the second derivative test to classify these critical points:

Find the second derivative of \( f(x) \):

\(f''(x) = \frac{d}{dx}(3x^2 + 30x - 33) = 6x + 30\)

Evaluate the second derivative at each critical point:

• At \(x = 1\): \(f''(1) = 6(1) + 30 = 36\). Since \(f''(1) > 0\), there is a local minimum at \( x = 1 \).
• At \(x = -11\): \(f''(-11) = 6(-11) + 30 = -36\). Since \(f''(-11) < 0\), there is a local maximum at \( x = -11 \).

Conclusion:

The function \( f(x) \) has a local maximum at \( x = -11 \). Therefore, the correct answer is: \(f(x)\) has a local maximum.