Question

Find whether the following function is differentiable at x = 1 and x = 2 or not : \[f\left( x \right) = \begin{cases}x, & & x < 1 \\ 2 - x, & & 1 \leq x \leq 2 \\ - 2 + 3x - x^2 , & & x > 2\end{cases}\] .

Answer 1

We have \[f\left( x \right) = \begin{cases}x, & x < 1 \\ 2 - x, & 1 \leq x \leq 2 \\ - 2 + 3x - x^2 , & x > 2\end{cases}\]

Clearly, f(x), being a polynomial function, is continuous and differentiable for all x < 1, 1 < x < 2 and also for all x > 2.

Thus, the possible points of non-differentiability of f(x) are x = 1 and x = 2.

Now,

f(1) = 2 – 1 = 1

and

f(2) = 2 – 2 = 0

At x = 1,

\[\begin{array}{cl}LHD & = & \lim_{x \to 1^-} \frac{f\left( x \right) - f\left( 1 \right)}{x - 1} \\ = & \lim_{x \to 1} \frac{x - 1}{x - 1} \\ = & 1 \\ RHD & = & \lim_{x \to 1^+} \frac{f\left( x \right) - f\left( 1 \right)}{x - 1} \\ = & \lim_{x \to 1} \frac{\left( 2 - x \right) - 1}{x - 1} \\ = & \lim_{x \to 1} \frac{- \left( x - 1 \right)}{x - 1} \\ = & - 1\end{array}\]

∴ LHD ≠ RHD

So, f(x) is not differentiable at x = 1.

At x = 2,

LHD =`lim_(x →2^-) (f(x) - f (2))/(x-2)`

        = `lim_(x →2) (2- x- 0)/(x-2)`

        = `lim_(x →2) (-(x-2))/(x-2)`

         = -1

RHD = `lim_(x →2^+) (f(x) - f (2))/(x-2)`

         = `lim_(x →2) ((-2+3x -x^2)-0)/(x-2)`

          = `lim_(x →2) (-(x^2- 3x +2))/(x-2)`

          = `lim_(x →2) (-(x-1)(x-2))/(x-2)`

           =` lim_(x →2) -(x-1)`

= -1

LHD = RHD

So, f(x) is differentiable at x = 2.

Thus, the given function is differentiable at x = 2, but not at x = 1.