Question

Let f : R $\to$ R be a function defined by $f(x)=max\{x,x^3\}$. The set of all points where $f(x)$ is NOT differentiable is

• A. {-1, 1}
• B. {-1, 0}
• C. {0, 1}
• D. {-1, 0, 1}

Answer 1

Answer: (D)

$f\left(x\right)=\max \left\{x,x^3\right\}=\{\begin{array}{ll}x& ;x<-1\\ x^3& ;-1\le x\le 0\\ x& ;0\le x\le 1\\ x^3& ;x\ge 1\end{array}$
$\therefore f^{\prime }(x)=\{\begin{array}{ll}1& ;x<-1\\ 3x^2& ;-1\le x\le 0\\ 1& ;0\le x\le 1\\ 3x^2& ;x\ge 1\end{array}$
Clearly f is not differentiable at - 1, 0 and 1.
ALTERNATE SOLUTION KEY CONCEPT
A continuous function f (x) is not differentiable at x = a
if graphically it takes a sharp turn at x = a.
Graph of f (x) = max $\{x,x^3\}$ is as shown with solid lines.
From graph of f (x) at x = - 1, 0, 1, we have sharp turns.
$\therefore$ f (x) is not differentiable at x = - 1, 0, 1.