Hi friends
how i solve it!
find the value of ( a ) so that given function f(x) is continous.
when f(x) is x^2+a if -2 < x < 2
6at if x=2
x-2 if x>2
please help me
I assume that you mean:
$\displaystyle f(x)=\left\{\begin{array}{rcl}x^2+a&~if~&-2<x<2 \\ 6at&~if~& x=2 \\ x-2&~if~&x>2\end{array}\right.$
1. Since $\displaystyle \lim_{x\to 2^+}f(x) = 0$ you can determine a because $\displaystyle \lim_{x\to2^-}f(x) $ must be zero too.
2. Now calculate t such that $\displaystyle f(2)= \lim_{x\to2^-}f(x) = \lim_{x\to 2^+}f(x)$