Math CentralQuandaries & Queries


Is f(x)=[(x^2-1)/(x-1) and x=2 at x=1] differentiable at x=1 ? Why ?

First of all $x^2 - 1 = (x - 1)(x + 1).$

When $x \ne 1$ the denominator of $f(x)$ is not zero so $f(x)$ simplifies to $x + 1$ which is differentiable.

When $x = 1$ the denominator of $f(x)$ is zero so $f(1)$ is not defined. Thus $f(x)$ is not differentiable at $x = 1.$


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