Question

Let $f: R \rightarrow R$ be defined as, $f(x) = \begin{cases} -55x, & \text{if } x < -5 \\ 2x^3 -3x^2 -120x, & \text{if } 5 \le x \le 4 \\ 2x^3 -3x^2-36x-336& \text{if } x > 4, \end{cases} $ Let $A=\{ x \in R : f$ is increasing $\} .$ Then $A$ is equal to

• A. $(-\infty,-5) \cup(4, \infty)$
• B. $(-5, \infty)$
• C. $(-\infty,-5) \cup(-4, \infty)$
• D. (-5,-4)$\cup(4, \infty)$

Answer 1

The Correct Option is D

 To determine where the function \(f\) is increasing, we need to analyze each piece of the piecewise function separately by considering its derivative.

The given function is:

• If \(x < -5\), then \(f(x) = -55x\).
• If \(-5 \le x \le 4\), then \(f(x) = 2x^3 - 3x^2 - 120x\).
• If \(x > 4\), then \(f(x) = 2x^3 - 3x^2 - 36x - 336\).

We differentiate each part:

1. For \(x < -5\), \(\frac{d}{dx}(-55x) = -55\). Since this is negative, \(f(x)\) is not increasing for \(x < -5\).
2. For \(-5 \le x \le 4\), compute the derivative: 

\[f'(x) = \frac{d}{dx}(2x^3 - 3x^2 - 120x) = 6x^2 - 6x - 120.\]

1. Solving \(6x^2 - 6x - 120 = 0\) gives the critical points: 

\[x^2 - x - 20 = 0 \quad \Rightarrow \quad (x - 5)(x + 4) = 0.\]

1. Thus, \(x = 5\) and \(x = -4\). For interval testing, between \(-5\) to \(4\), pick test points to determine the sign of \(f'(x)\):
   • For \(x = -4.5\) (test point in \((-5, -4)\)), \(f'(-4.5) > 0\).
2. For \(x > 4\), compute the derivative: 

\[f'(x) = \frac{d}{dx}(2x^3 - 3x^2 - 36x - 336) = 6x^2 - 6x - 36.\]

1. Factoring gives: 

\[f'(x) = 6(x^2 - x - 6) = 6(x - 3)(x + 2).\]

1. Since we are interested in \(x > 4\), observe sign changes or test points:
   • For large \(x = 5\) (in \((4, \infty)\)), \(f'(5) > 0\), indicating increasing.

Thus, the function \(f(x)\) is increasing in the intervals \(-5, -4) \cup (4, \infty)\).