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.
The answer: 30 .
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Explanation:
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10 + (x/6 ) = 5 ; solve for "x" ;
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Subtract "10" from each side ;
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10 + (x/6 ) - 10 = 5 - 10 ;
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to get:
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x / 6 = -5 ;
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Now, multiply EACH side of the equation by "6" ; to isolate "x" on one side of the equation; and to solve for "x" ;
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(x/6) *6 = -5 * (6) ;
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To get: x = 30 .
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The answer is 0.
1 • -2 = -2
-2 + -2 = -4
-4 • -2 = 8
-8 + 8 = 0
0 • -2 = 0