Continuity of f(x)=piecewise function

[tylersmith7690]

Homework Statement

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

Homework Equations

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

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.

thanks in advance.

 

[vela]

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?

 

[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.

 

[vela]

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

 

[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?

 

[vela]

Yes, do what I've already suggested twice.

 

[pasmith]

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?

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

 

[Evo]

Homework Statement

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

Homework Equations

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

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.

thanks in advance.

Quoted for the record.

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