Question

Let $f:R\to R$ be defined as:

$$\begin{gathered} f\left(x\right)=\left\{-55x,ifx<-5 \\ 2x^3-3x^2-120x,if-5\le x\le 4 \\ 2x^3-3x^2-36x-336,ifx>4\right. \end{gathered}$$

Let $A=\left\{x\in R:f\text{is increasing}\right\}$. Then $A$ is equal to:

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

Answer 1

The correct option is A

$\left(-5,-4\right)\cup (4,\infty )$

Determine the value of $A$

Step 1; Determine $f\text{'}\left(x\right)$

If we differentiate $f\left(x\right)$ and we get $f\text{'}\left(x\right)$ as positive, it means that the function is increasing with the increase in the value of $x$.

Differentiating $f\left(x\right)$:

$$\begin{gathered} f\text{'}\left(x\right)=\left\{-55,ifx<-5 \\ 6x^2-6x-120,if-5\le x\le 4 \\ 6x^2-6x-36,ifx>4\right. \end{gathered}$$

$$\begin{gathered} f\text{'}\left(x\right)=\left\{-55,ifx<-5 \\ 6(x^2-x-20),if-5\le x\le 4 \\ 6(x^2-x-6),ifx>4\right. \end{gathered}$$

$$\begin{gathered} f\text{'}\left(x\right)=\left\{-55,ifx<-5 \\ 6(x-5)(x+4),if-5\le x\le 4 \\ 6(x-3)(x+2),ifx>4\right. \end{gathered}$$

We can observe, that $f\text{'}\left(x\right)$ is negative when $x<-5$.

Step 2: Determine the critical point when $-5\le x\le 4$:

$$\begin{gathered} \Rightarrow (x-5)(x+4)=0 \\ \Rightarrow x=5,-4 \end{gathered}$$

So, $f\text{'}\left(x\right)$ is positive at $(-\infty ,-4)$ and $(5,\infty )$, but our domain says $-5\le x\le 4$.

Hence,$f\text{'}\left(x\right)$ is positive at $(-5,-4)$

Similarly,

$$\begin{gathered} \mathrm{for}x>4, \\ \Rightarrow 6(x-3)(x+2)=0 \\ \Rightarrow x=3,-2 \end{gathered}$$

So, $f\text{'}\left(x\right)$ is positive at $(-\infty ,-2)$ and $(3,\infty )$, but the domain says: $x>4$

Therefore, $f\text{'}\left(x\right)$ is positive at $(4,\infty ).$

Hence, the value will increase at $(-5,-4)\cup (4,\infty ).$

Therefore, the correct answer is Option (A).