Question

Describe the \(x\) -values at which \(f\) is differentiable. $$ f(x)=\frac{1}{x-1} $$

Short answer

The function \(f(x) = \frac{1}{x-1}\) is differentiable for all real numbers except \(x=1\).

Step-by-step solution

Identify any points of discontinuity

Identify points where the function is not defined to start with. These points will be discontinuities of the function. In this case, \(f(x)\) is not defined when the denominator of the fraction is zero i.e., \(x=1\).

Find the derivative

Next, compute the derivative of the given function. The derivative of \(f(x) = \frac{1}{x-1}\) is \(f'(x) = -\frac{1}{{(x-1)^2}}\).

Identify any points at which the derivative does not exist

Now that the derivative has been found, identify any point at which the derivative does not exist. Here, the derivative is not defined at \(x=1\). Thus, the function \(f\) is not differentiable at \(x=1\).

Identify the values of \(x\) at which \(f(x)\) is differentiable

By eliminating the \(x\)-values where \(f(x)\) is not differentiable, specify the values of \(x\) at which \(f(x)\) is differentiable. In this case, \(f\) is differentiable for \(x \neq 1\). Hence, \(f\) is differentiable for all real values of \(x\) except 1.