Maths · Limits, continuity and differentiability
=\left\{\begin{array}{cc}-1, & -2 \leq x<0 \\ x^{2}-1, & 0<x \leq 2\end{array} \
\( \operatorname{Let} f(x)=\left\{\begin{array}{cc}-1, & -2 \leq x<0 \\ x^{2}-1, & 0<x \leq 2\end{array} \) and \right. \( \boldsymbol{g}(\boldsymbol{x})=|\boldsymbol{f}(\boldsymbol{x})|+\boldsymbol{f}|\boldsymbol{x}| \) then the number of points which \( g(x) \) is non differentiable, is
- A. at most one point
- B. 2
- C. exactly one point
- D. infinite
Step-by-step solution
The function g(x) = |f(x)| + f(|x|) simplifies piecewise: for x in [-2,0), g(x)=x^2; for (0,1], g(x)=0; for (1,2], g(x)=2x^2-2. At x=1, left derivative is 0 and right derivative is 4, so g is non-differentiable at x=1. No other points cause non-differentiability. Hence exactly one point.
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