# 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.