Let f(x)={−1,x2−1,−2≤x<00≤x≤2 and g(x)=∣f(x)∣+f(∣x∣). Then, in the interval (−2,2), g is:
Solutions · 1
Official solution
f(∣x∣)=x2−1 for −2≤x≤2.
\[g(x)={−1+x2−1,(x2−1)+∣x2−1∣,−2≤x<00≤x≤2=⎩⎨⎧x2−2,0,2(x2−1),−2≤x<00≤x<11≤x≤2\]
Checking derivatives at x=0 (both 0) and x=1 (LHD =0, RHD =4) shows g is non-differentiable at exactly one point, x=1.