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in Continuity and Differentiability by (91.5k points)
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Show that the function `f` defined as follows `f(x)={3x-2 , 0ltxle1 ; 2x^2-x , 1ltxle2 ; 5x-4 , xgt2,` is continous at x=2 but not differentiable.

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`f(x)=f(x)=f(x^2)`
`f(2^-)=lim_(h->0)(2(2-h)^2)-(2-h)`
`=lim_(h->0)(2(4-4h+h^2)-2+h`
`=lim_(h->0)(6-7h+2h^2)=6`
`f(2^+)=lim_(h->o)(5(2+h)-4`
`=10+5h-4=6`
`f(x^-)=f(x)=f(x^2)`
f(x) is continuous at x=2
LHD`=lim_(h->0)(f(2-h)-f(2))/(-h)`
`=lim_(h->0)(6-7h+2h^2-6)/(-h)`
`=7`
RHD`lim_(h->0)(f(2+h)-f(2))/h`
`=lim_(h->0)(6+5h-6)/h`
`=5`
LHD`!=`RHD
f(x) is not different at x=2.

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