Question

The function f(x) is defined as follows f(x) = $\{\begin{array}{cc}-x& \text{if}x<0\\ x^2& \text{if}0\le x\le 1thenf\left(x\right)is\\ x^3-x+1& \text{if}x>1\end{array}$

• A. derivable & continuous at $x=0$
• B. derivable at $x=1$ but not continuous at $x=1$
• C. neither derivable nor continuous at $x=1$
• D. not derivable at $x=0$ but continuous at $x=1$

Answer 1

Answer: (D)

not derivable at $x=0$ but continuous at $x=1$

f(x) = $\{\begin{array}{cc}-x& \text{if}x<0\\ x^2& \text{if}0\le x\le 1\\ x^3-x+1& \text{if}x>1\end{array}$
Clearly $f\left(x\right)$ is continuous $\forall x$
f'(x) = $\{\begin{array}{cc}-1& \text{if}x<0\\ 2x& \text{if}0\le x\le 1\\ 3x^2-1& \text{if}x>1\end{array}$
Also $f\left(x\right)$ is not differentiable at $x=0$ and differentiable at $x=1$