Question

Determine all numbers at which the function is continuous.

Answer 1

Step-by-step explanation:

If the graph of any function is an unbroken curve, then the function is continuous. Let's study the function at the the point [/tex]x=5[/tex]:

At this point the function has the following value:

$f(5)=-\frac{3}{4}$, so the function in fact exists here, but let's find the limit here using:

$f(x)=\frac{x^2-7x+10}{x^2-14x+45}$

So:

$\underset{x\rightarrow5}{lim}\frac{x^2-7x+10}{x^2-14x+45}$

By factoring out this function we have:

$\underset{x\rightarrow5}{lim}\frac{(x-2)(x-5)}{(x-5)(x-9)} \\ \\ \therefore \underset{x\rightarrow5}{lim}\frac{(x-2)}{(x-9)} \\ \\ \therefore \frac{(5-2)}{(5-9)}=-\frac{3}{4}$

Since $\underset{x\rightarrow5}{lim}f(x)=f(5)$ then the function is continuous here.

Let's come back to our function:

$f(x)=\frac{x^2-7x+10}{x^2-14x+45}$

If we factor out this function we get:

$f(x)=\frac{(x-2)}{(x-9)}$

Notice that at x = 9 the denominator becomes 0 implying that at this x-value there is a vertical asymptote. The graph of this function is shown below and you can see that at x = 9 the function is not continous

Therefore, the answer is:

b. continous at every point exept $x=9$