Determine the values of x for which the following function fails to be continuous or differentiable: `f(x)={1-x, x < 1 (1-x)(2-x),1 <= x <= 2 3-x, x > 2` justify your answer. `

Answer

`(1, 2)`

Explanation

It is clear thet the given function
`f(x) = {{:((1-x)", "x lt 1 ),((1-x)(2-x)", "1 le x le 2 ),((3 - x)", " x gt 2 ):}`
continuous and differentiable at all points except possibly at x = 1 and 2.
Continuity at x = 1 ,
LHL ` = underset( x to 1^(-)) lim f (x) = underset ( x to 1^(-)) lim (1-x)`
` = underset( h to 0) lim [1-(1-h)] = underset( h to 0) lim h = 0 `
and RHL = ` underset( x to 1^(+)) lim f(x) = underset( x to 1^(+)) lim (1-x) (2-x) `
`= underset( h to 0) lim [1-h)][2-(1+h)]`
` = underset( h to 0) lim -h*(1-h) =0 `
`:. ` LHL = RHL = f(1) = 0
Therefore, f is continuous at x = 1
Differentiability at x = 1,
`L f' (1) = underset( h to 0) lim (f(1-h)-f(1)) / (-h) `
` = underset( h to 0) lim (1-(1-h)-0) / (-h) = underset( h to 0) lim (h/(-h)) =- 1 `
and ` R f'(1) = underset( h to 0) lim (f(1+h)-f(1))/h `
`= underset( h to 0) lim ([1-(1+h)][(2-(1+h)]-0))/h `
` = underset( h to 0) lim (-h(1-h))/h = underset( h to 0) lim (h-1) =- 1`
Since, ` L[f'(1)]= Rf'(1)`, therefore f is differentiable at x = 1.
Continuity at x = 2,
LHL `= underset( x to 2^(-))lim f (x) = underset( x to 2^(-)) lim (1-x)(2-x) `
` = underset( h to 0) lim [1-h)][(2-(2-h)]`
`underset( h to 0) lim (-1+h) h = 0 `
and RHL ` = underset( x to 2^(+)) lim f(x) = underset( x to 2^(+)) lim (3-x) `
` = underset( h to 0) lim [3-(2+h)]= underset( h to 0) lim (1-h) = 1 `
Since, LHL ` ne ` RHL , therefore f is not continuous at x = 2 as such f cannot be differentiable at x = 2 .
Hence, f is continuous and differentiable at all points except at c = 2.