Show that the function `f(x)={|2x-3|[x]xlt=0` and `sin((pix)/2)`,`x>0` is continuous but not differentiable at `x=0`

Explanation

Let`f(1)=∣2−3∣[1]=1−(i)`

So,`lim_(h->0)f(1+h)=∣2+2h−3∣[1+h]=1×1=1−(ii)`

Also,`lim_(h->0)f(1−h)=sin((pi(1−h))/2)=sin(pi/2)=1−(iii)`

Hence f(x) is continuous at 1

For differentiability `lim_(h->0)(f(1+h)−f(1))/h `must be equal to `lim_(h->0)(f(1)−f(1−h))/h`

=`lim_(h->0)(∣2+2h−3∣[1+h]−∣2−3∣[1])/h= lim_(h->0)(∣2−3∣[1]−sin((pi−h)/2))/h`

=` lim_(h->0)(∣2h−1∣−1)/h= lim_(h->0)(1−sin((pi−h)/2))/h`

Since h is very small, (2h−1) is−ve.

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