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  • Let <img alt="f(x)=\left\{\begin{matrix} max \left\{\left|x\right|,\:x^2\right\}, &|x|\leq 2 \\ 8-2|x| & 2<|x|\leq 4 \end{matrix}\right." src="https://learn.careers360.com/latex-image/?f%28x%29%3D%5Cleft%5C%7B%5Cbegin%7Bmatrix%7D%20max%

Let f(x)=\left\{\begin{matrix} max \left\{\left|x\right|,\:x^2\right\}, &|x|\leq 2 \\ 8-2|x| & 2<|x|\leq 4 \end{matrix}\right. Let S be the set of points in the interval (-4,4) at which f is not differentiable. Then S:


Option: 1 is an empty set
Option: 2 equals {-2,-1,0,1,2}
Option: 3 equals{-2,-1,1,2}
Option: 4 equals{-2,2}
 

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best_answer

 

Condition for differentiability -

A function  f(x) is said to be differentiable at  x=x_{\circ }  if   Rf'(x_{\circ })\:\:and\:\:Lf'(x_{\circ })   both exist and are equal otherwise non differentiable

-

 

 

 

Geometrical interpretation of Derivative -

Let P be any point (x, y)  on the curve  y = f(x) and Q is a point in the neighbourhood of  P  on either side of  P. such that the co-ordinate of the point Q are 

(x+\delta x, y+\delta y)  satisfying the curve y = f(x)

\therefore \:M_{T}=slope\:of\:tangent

=\lim_{\delta x\rightarrow \circ }\:\:\frac{(y+\delta y)-y}{(x+\delta x)-x}=\lim_{\delta x\rightarrow \circ }\:\:\frac{\delta y}{\delta x}


\therefore \:\:M_{T}=(\frac{dy}{dx})\:\:at \;\:(x,\:y)

- wherein


Where (x, y)  on the curve and  MT  is tangent at (x, y).

 

f(x) = \left\{\begin{matrix} 8+2x, & -4\leq x<-2 \\ x^{2} ,&-2\leq x\leq -1 \\ \left | x \right |, &-1<x<1 \\ x^{2}, & 1\leq x\leq 2\\ 8-2x, & 2< x\leq 4 \end{matrix}\right.

Plot graph

f(x) is not differentiable at

x = -2,-1,0,1,2.

Posted by

Ritika Jonwal

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