Question

Let the functions $f:R\to R$ and $g:R\to R$ be defined as:

$f\left(x\right)=\left\{\begin{array}{ll}x+2& x<0\\ x^2& x⩾0\end{array}\right.$and $g\left(x\right)=\left\{\begin{array}{ll}{x}^3& x<1\\ 3x-2& x⩾1\end{array}\right.$

Then, the number of points in $R$ where $(f\circ g)\left(x\right)$ is NOT differentiable is equal to:

• A. $1$
• B. $2$
• C. $3$
• D. $0$

Answer 1

The correct option is A

$1$

Explanation for the correct option:

The functions $f:R\to R$ is defined as $f\left(x\right)=\left\{\begin{array}{ll}x+2& x\le 0\\ x^2& x⩾0\end{array}\right.$ and the function $g:R\to R$ is defined as $g\left(x\right)=\left\{\begin{array}{ll}{x}^3& x<1\\ 3x-2& x⩾1\end{array}\right.$.

To find:

The number of points of non-differentiability.

Explanation:

Now finding the number of points of non-differentiability, we get

$f\circ g\left(x\right)=\left\{\begin{array}{l}{x}^3+2x<0\\ x^{6}0\le x<1\\ {(3x-2)}^{2}x\ge 1\end{array}\right.$

Clearly $f\circ g\left(x\right)$is discontinuous at$x=0$ then non-differentiable at $x=0$

Using $f\left(x\right)=\frac{f\left(x+h\right)-f\left(x\right)}{x-h}$

Now, at $x=1$

$\begin{array}{rcl}f\left(1^{+}\right)& =& \underset{h\to 0+}{\lim }\left[\frac{\left[f\right(1+h)–f(1\left)\right]}{h}\right]\\ & =& \underset{h\to 0^{+}}{\lim }\left[\frac{{\left(3\left(1+h\right)-2\right)}^2-1}{h}\right]\\ & =& 6\end{array}$

$\begin{array}{rcl}f\left(1^{-}\right)& =& \underset{h\to 0^{-}}{\lim }\left[\frac{\left[f\right(1-h)–f(1\left)\right]}{\left[h\right]}\right]\\ & =& \underset{h\to 0^{-}}{\lim }\left[\frac{{\left(1-h\right)}^6-1}{-h}\right]\{L\text{'}HospitalRule\}\\ & =& \underset{h\to 0^{-}}{\lim }\left[\frac{-6{\left(1-h\right)}^5}{-1}\right]\\ & =& 6\end{array}$

$\Rightarrow f\left(1^{+}\right)=f\left(1^{-}\right)$

Therefore, the number of points of non-differentiability $=1$.

Hence, the correct answer is option (A).