# Let f(x)= frac{x^2+4x+3}{x+1} A student recognizes that this function can be simplified as...

## Question:

Let {eq}f(x)= \frac{x^2+4x+3}{x+1} {/eq}

A student recognizes that this function can be simplified as follows:

{eq}f(x)=\frac{x^2+4x+3}{x+1} = \frac{(x+1)(x+3)}{x+1} = x+3 {/eq}

Since {eq}y = x+3 {/eq} is a line with slope 1, the student makes the following conclusions:

{eq}f'(-2)=1 ;\; f'(-1)=1 ;\; f'(0)=1 ; \;f'(1)=1 {/eq}.

Where did the student make an error?

## Discontinuities of Rational Functions:

This sort of problem illustrates the idea of discontinuities of rational functions. Specifically, a rational function is not differentiable at a discontinuity. Input values that lead to a denominator of zero are a common type of discontinuity seen in rational functions.

The original function {eq}f(x)= \dfrac{x^2+4x+3}{x+1} {/eq} is not defined when {eq}x = -1 {/eq}

(because that value of x leads to the indeterminate form {eq}\frac00 {/eq}).

Therefore, the original function has a discontinuity at that point.

So, the cancellation from {eq}\dfrac{x^2+4x+3}{x+1} {/eq} to {eq}x + 3 {/eq} is true only if {eq}x \neq -1. {/eq}

Hence, the function is not differentiable at {eq}x = -1 {/eq}, and so the conclusion that {eq}f'(-1) = 1 {/eq} is invalid.