## anonymous 5 years ago Check if the function is continuous.

1. anonymous

$f(x) = \left\{ {{{x^2 + 6x - 16} \over {x^2 + x -6}}, x \neq 2} \right\}$ ${7x - 4}, x =2$

2. anonymous

I can't order the brackets

3. myininaya

x^2+x-6=(x+3)(x-2) x^2+6x-16 doesn't have either factor so the limit does not exsit at -3 and 2 so the function is not continuous at either x

4. myininaya

oh x=2 is defined for 7x-4

5. myininaya

still since the limit doesn't exist at x=2 then f is not continuous there

6. anonymous

yeap. sorry for that.

7. anonymous

Oh! OK! Thanks.

8. myininaya

all you have to do is make sure the limit exists and the limx->af(x)=f(a) then f is continuous at x=a

9. anonymous

$\lim_{x\rightarrow 2}f(x)=\lim_{x\rightarrow 2}\frac{x^2+6x-16}{x^2+x-6}=\lim_{x\rightarrow 2}\frac{(x+8)(x-2)}{(x+3)(x-2)}=\lim_{x\rightarrow 2}\frac{x+8}{x+3}=2$ $f(2)=7\cdot 2-4=10$ $\lim_{x\rightarrow 2}f(x)\neq f(2)\Rightarrow f(x)\quad not\enspace continuous$

10. anonymous

Thanks!