Continuity question

1. Aug 17, 2013

tylersmith7690

1. The problem statement, all variables and given/known data

For which values of a E ℝ, is the function given by

f(x) = pieceise function
x^2+4x-4, x<a
cos((x-a)/2) , x ≥ a.

continuous at x=a

2. Relevant equations

I'm getting stuck on the algebra part to be honest.

3. The attempt at a solution
lim x→a f(x)= f(a) to be continuous

lim x→a- = x^2+4x-4 and this must be equal to lim x→a+ = cos((x-a)/2)

I think its 1 because cos(1-1/2)=1 and 1+4-4=1

I'm just confused on the working out part, how to I algebraically manipulate the equations to show this? or am I completely wrong.

2. Aug 17, 2013

vela

Staff Emeritus
You seem to have the basic idea. You need
$$\lim_{x \to a^-} x^2 + 4x - 4 = \lim_{x \to a^+} \cos\frac{x-a}{2}.$$ So what are the two sides equal to?

3. Aug 17, 2013

tylersmith7690

Thanks for the reply, I'm guessing the value of (a) in the equation has to be 1 for both sides of the equation to equal 1.
I'm almost certain a=1 for the function to be continuous, just having a hard time showing the working out besides me just putting a=1 into the limit equation.

Thanks again.

4. Aug 17, 2013

vela

Staff Emeritus
Don't plug any value in for a. Just work out what the limits equal in terms of a.

5. Aug 17, 2013

tylersmith7690

Thank you for your replies, but I literally have no idea of how to solve a, i think its the cosine function that is throwing me off. Any tips for trying to solve it?

6. Aug 17, 2013

vela

Staff Emeritus
Yes, do what I've already suggested twice.

7. Aug 17, 2013

pasmith

You are given that if $x \geq a$ then $f(x) = \cos((x - a)/2)$. Therefore $f(a) = \cos((a-a)/2) = \cos(0) = 1$.

8. Aug 17, 2013

Staff: Mentor

Quoted for the record.

Please remember to ALWAYS quote the OP's posts.