$f (x) = 2 ∣ ∣ \frac{x - 1}{x - 1} ∣ ∣ \forall x & x \neq 1$ , What should be the value of f(x) at x=1 so that it is differentiable from $(- \infty, \infty)$
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Solution

(A) Lets look at the graph of f(x), f(x) is discontinuous at x=1

For $x ϵ (- \infty, 1) \cup (1, \infty) f (x) = 2$

At x=1,Left hand Derivative $= lim △ x \to 0^{+} \frac{2 ∣ ∣ \frac{x + △ x - 1}{x + △ x - 1} ∣ ∣ - 2 ∣ ∣ \frac{x - 1}{x - 1} ∣ ∣}{△ x}$

$= 0$

Right hand Derivative $= lim △ x \to 0^{-} \frac{2 ∣ ∣ \frac{x + △ x - 1}{x + △ x - 1} ∣ ∣ - 2 ∣ ∣ \frac{x - 1}{x - 1} ∣ ∣}{△ x}$

$= 0$

$lim x \to 0^{+} f (x) = 2, lim x \to 0^{-} f (x) = 2,$

We need f(x) to be continuous for it be differentiable.

So value of f(x) at x=1 such that its continuous is 2.

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