Given, f(x)‌‌=|2x+1|−3|x+2|+|x2+x−2| ‌‌=|2x+1|−3|x+2|+|x+2|×|x−1| Here, critical points are x=‌
−1
2
,−2,1 ∴f(x)={
x2+2x+3
‌
‌
x<−2
−x2−6x−5
‌
‌
−2<x<‌
−1
2
−x2−2x−3
‌
‌
‌
−1
2
<x<1
x2−7
‌
‌
x>1.
Now, f′(x)={
2x+2
‌
‌
x<−2
−2x−6
‌
‌
−2<x<‌
−1
2
−2x−2
‌
‌
−1
2
<x<1
2x‌‌
x>1.
Now, f′(x) at 1,−2 and −1∕2 are For x=1, f′(x)‌‌=2x=2×1=2 ‌ and ‌−2x−2‌‌=−(2×1)−2=−4 f′(x)=2x= and −2x−2=−(2 both are not equal. ∴ Non-differentiable at x=1 Similarly, for x=−2,f′(x)=2x+2=2×(−2)+2=−2 and −2x−6=−2×(−2)−6=−2 both are equal. ∴ Differentiable at x=−2 and for x=−1∕2,f′(x)=−2x−6 =−2×(‌
−1
2
)−6=−5‌ and ‌ −2x−2=−2×(‌
−1
2
)−2=−1 both are not equal. ∴ Non-differentiable at x=−1∕2 ∴ The number of points at which f(x) is non-differentiable is 2 .