Show that the function `f(x)={|2x-3|\ [x],\ \ \ xgeq1sin((pix)/2),\ \ \ \ \ \ x<1` is continuous but not differentiable at `x=1` .

Explanation

`f(1)=∣2−3∣[1]=1`
`lim_(h->0) f(1+h) = |2+2h-3|[1+h]`
`=1 xx 1`
`=1`
`lim_(h->0)f(1-h) `
`=sin((pi(1-h))/2)`
`=sin (pi/2)`
`=1`
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= (|2-3|[1]-sin((pi-h)/2))/h`
`=>(|2h-1|-1)/h=(1-sin((pi-h)/2))/h[0/0 form ]`
Since h is very
small,
`(2h−1)` is `−ve`.
`=>(1-2h-1)/h =-cos ((pi-h)/2) xx (-1/2) `
`=> -2 ne 0`
that means function is non-differentiable at 1.
but continuous at 1.
Hence proved.