Let f be a composite function of x defined by f(u)=1u^2+u-2, u(x)=1x-1.

Question

Let $f$ be a composite function of $x$ defined by $f(u)=\dfrac{1}{u^2+u-2}$, $u(x)=\dfrac{1}{x-1}$. Then the number of points $x$ where $f$ is discontinuous is:

Accepted Answer

$u(x)$ is discontinuous at $x=1$. $f(u)=\dfrac{1}{(u+2)(u-1)}$ is discontinuous at $u=-2,1$.

$u=-2\Rightarrow\dfrac{1}{x-1}=-2\Rightarrow x=\dfrac12$; $u=1\Rightarrow\dfrac{1}{x-1}=1\Rightarrow x=2$.

So the composite is discontinuous at $x=1,\dfrac12,2$ --- three points.

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