The number of points,at which the function f( | vedclass

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(C) Given function: $f(x) = |2x+1| - 3|x+2| + |x^2+x-2|$Factorize the quadratic term: $|x^2+x-2| = |(x+2)(x-1)| = |x+2||x-1|$Substitute back into the

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$2$

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(C) Given function: $f(x) = |2x+1| - 3|x+2| + |x^2+x-2|$Factorize the quadratic term: $|x^2+x-2| = |(x+2)(x-1)| = |x+2||x-1|$Substitute back into the function: $f(x) = |2x+1| - 3|x+2| + |x+2||x-1|$$f(x) = |2x+1| + |x+2|(|x-1| - 3)$The points where the expression inside the absolute value becomes zero are $x = -1/2$,$x = -2$,and $x = 1$.Let $g(x) = |x-1| - 3$. The function $f(x)$ is non-differentiable at points where the argument of an absolute value is zero,provided the function does not have a smooth turning point there.$1$. At $x = -1/2$,$|2x+1|$ is non-differentiable,and the other terms are smooth. Thus,$x = -1/2$ is a point of non-differentiability.$2$. At $x = 1$,$|x-1|$ is non-differentiable,and $|x+2| = 3 
eq 0$. Thus,$x = 1$ is a point of non-differentiability.$3$. At $x = -2$,we check the behavior: $f(x) = |2x+1| + |x+2|(|x-1|-3)$. Near $x = -2$,$|x-1| = -(x-1) = 1-x$. So $f(x) \approx |2x+1| + |x+2|(1-x-3) = |2x+1| + |x+2|(-x-2) = |2x+1| - |x+2|(x+2) = |2x+1| - (x+2)^2$. Since $(x+2)^2$ is differentiable at $x = -2$,the non-differentiability of $|x+2|$ is cancelled out by the factor $(x+2)$.Therefore,the points of non-differentiability are $x = -1/2$ and $x = 1$.The number of points of non-differentiability is $2$.

The number of points,at which the function $f(x) = |2x+1| - 3|x+2| + |x^2+x-2|$,$x \in R$ is not differentiable,is ............

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