Continuity of f(x)=piecewise function

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Homework Help Overview

The discussion revolves around determining the values of \( a \in \mathbb{R} \) for which a piecewise function is continuous at \( x = a \). The function is defined as \( f(x) = x^2 + 4x - 4 \) for \( x < a \) and \( f(x) = \cos\left(\frac{x-a}{2}\right) \) for \( x \geq a \).

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss the limits from both sides of \( a \) and the conditions for continuity. There is confusion regarding the algebraic manipulation required to equate the limits and the function value at \( a \). Some participants express uncertainty about how to approach the cosine function in this context.

Discussion Status

There is an ongoing exploration of the limits as \( x \) approaches \( a \) from both sides. Some participants have suggested specific values for \( a \) but have been advised to avoid plugging in values prematurely. The conversation indicates a lack of consensus on the algebraic steps needed to demonstrate continuity.

Contextual Notes

Participants mention difficulties with the algebra involved and the implications of the piecewise definition on continuity. There is a repeated emphasis on understanding the limits rather than substituting values directly.

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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.
 
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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?
 
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.
 
Don't plug any value in for a. Just work out what the limits equal in terms of a.
 
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?
 
Yes, do what I've already suggested twice.
 
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tylersmith7690 said:
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].
 
tylersmith7690 said:

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.

Please remember to ALWAYS quote the OP's posts.
 

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