Question 453993
Find the values of a and b that make f continuous everywhere. 
f (x)= 
{ 
1. x^2&#8722;4x&#8722;2 if x<2 
2. ax2&#8722;bx+3 if 2&#8804;x<3 
3. 2x&#8722;a+b if x&#8805;3 } 
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Evaluate "1" at x = 2
f(2) = 2^2-4*2-2 = -6
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Evaluate "3" at x = 3
g(3) = 2(3)-a+b = 6-a+b
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So two of the points of "2" must be (2,-6) and (3,6-a+b)
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Substitute those coordinates into "2"
to solve for "a" and for "b".
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2. ax2&#8722;bx+3
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(2,-6) gives you 4a-2b+3 = -6
2a - b = -9
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(3,6-a+b) gives you a(9)-b(3)+3 = 6-a+b
10a - 4b = 3
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Solve the system:
2a - b = -9
10a -4b = 3
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Multiply thru 1st equation by 5:
10a - 5b = -45
10a - 4b =  3
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Subtract to get:
b = 48
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Solve for "a":
2a - b = -9
2a - 48 = -9
2a = 39
a = 39/2
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I'd suggest you check alll the arithmetic
here as mistakes can creep in.
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Cheers,
Stan H.