Answers:
b = -7
c = 8
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Step-by-step explanation
Let
- g(x) = x+1
- h(x) = x^2+bx+c
We have f(x) = g(x) when 1 < x < 7.
When |x-4| ≥ 3, then f(x) = h(x).
Let's solve |x-4| ≥ 3 for x
|x-4| ≥ 3
x-4 ≥ 3 or x-4 ≤ -3
x ≥ 3+4 or x ≤ -3+4
x ≥ 7 or x ≤ 1
x ≤ 1 or x ≥ 7
Therefore, f(x) = h(x) = x^2+bx+c when x ≤ 1 or x ≥ 7
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Focus on the endpoints of the first piece. Plug x = 1 into g(x)
g(x) = x+1
g(1) = 1+1
g(1) = 2
Do the same for h(x).
h(x) = x^2+bx+c
h(1) = 1^2+b*1+c
h(1) = 1+b+c
The function pieces g(x) and h(x) connect to each other at x = 1 when g(1) = h(1) is the case.
So this would mean:
g(1) = h(1)
2 = 1+b+c
c = 2-1-b
c = 1-b
We'll use this later on.
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Now plug x = 7 into g(x)
g(x) = x+1
g(7) = 7+1
g(7) = 8
Repeat for h(x)
h(x) = x^2+bx+c
h(7) = 7^2+b*7+c
h(7) = 49+7b+c
Plug in c = 1-b found earlier
h(7) = 49+7b+c
h(7) = 49+7b+(1-b)
h(7) = 6b+50
Like with the previous section, the functions g(x) and h(x) connect at x = 7 when g(7) = h(7) is true.
So,
g(7) = h(7)
8 = 6b+50
6b+50 = 8
6b = 8-50
6b = -42
b = -42/6
b = -7
We can then determine c.
c = 1-b
c = 1-(-7)
c = 1+7
c = 8
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In summary, b = -7 and c = 8 are the final answers.
I recommend using a tool like GeoGebra or Desmos to verify the answers.