Q. 20 - Calculus

Question

Graph the functions $f\left(x\right)=2-x$ and $g\left(x\right)=\frac{4-x^2}{x+2}$, and show that they are equal everywhere except at one point. Then show that f(x) and g(x) have different values, but the same limit, at this point. 

Step-by-Step Solution

Answer

Both functions approaches -1 as x approachs -2.

Step 1. Given information.

Given functions are:

$$\begin{gathered} f\left(x\right)=2-x \\ g\left(x\right)=\frac{4-x^2}{x+2} \end{gathered}$$

We have to graph the function.

Step 2. Graph the f(x)

Graph of $f\left(x\right)=2-x$ is:

Step 3. Graph the g(x)

Graph of $g\left(x\right)=\frac{4-x^2}{x+2}$ is:

Step 4. Compare the graphs.

The graph of g looks similar to the graph of f but with the hole at $x=-2$.

$f(-2)=-4$ and $g(-2)$ is undefined.

Both functions approach -4 as x approaches -2.