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I REALLY NEED HELP WITH MY CALC CORRECTIONS. THIS IS THE THIRD TIME IVE POSTED THIS. I NEED HELP ASAP PLS!! I HATE CALC 1

I REALLY NEED HELP WITH MY CALC CORRECTIONS. THIS IS THE THIRD TIME IVE POSTED THIS-example-1
User Willjay
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1 Answer

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1. As x approaches 1, f(x) approaches 3-1, that is 2.

If a = 2 and b = 3, f(1) = 2(1)^2 + 3(1) = 5

So there is a 'jump' in values of x at x = 1. So its not continuous at x=1.

2. For continuity at x = 1, ax^2 + bx must = 2 that is when a + b = 2.

3. for continuity at x = 2 , ax^2 + bx must be = 0 that is when a*2^2 + 2b = 0

- that is 4a + 2b = 0.

4. we have system of equations:-

a + b = 2
4a + 2b = 0

this gives a = -2 and b = 4 So f(x) is continuous when a = -2 and b = 4.
User PixelPaul
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