Posted by **Anonymous** on Sunday, September 19, 2010 at 12:00pm.

Consider the function f(x)=

x^2+a if x<0,

2x+a^2 if 0 less than or equal to x less than or equal to 1,

3 if x>1

a. determine the value(s) of a that will make f(x) continuous at x=0

b. determine the value or values of a that will make f(x) continuous at x=1

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**bobpursley**, Sunday, September 19, 2010 at 12:01pm
I will be happy to critique your thinking.

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**Anonymous**, Sunday, September 19, 2010 at 12:11pm
For part a I took the limit as x tends to zero from the right of 2x+a^2. I put zero in for x and then said a=0 which would also make the limit as x approaches zero from the left of x^2+a 0 so the function would be continuous.

I don't know how to start part b because there is no variable for x>1

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**bobpursley**, Sunday, September 19, 2010 at 12:13pm
You are correct, part of 3 is missing.

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**Anonymous**, Sunday, September 19, 2010 at 12:14pm
Maybe the limit of x approaches 1 from the right of 3 = 3. And then set the limit of x approaches 1 from the left of 2x+a^2 = 3 and then solve for a and get 1.

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**bobpursley**, Sunday, September 19, 2010 at 12:15pm
I am not certain I would make that assumption on the problem. However, you could be right.

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**Anonymous**, Sunday, September 19, 2010 at 12:17pm
one more question.

if possible determine the value(s) of a that will make f(x) continuous on (negative infinity, infinity)

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**MathMate**, Sunday, September 19, 2010 at 1:02pm
There is more than one value of a that will make the function continuous at x=0, since at x=0,

x^2+a =2x+a^2

or setting x=0,

a-a²=0

a(1-a)=0

giving

a=0 or a=1.

Now you can continue with the last part.

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