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November 12th, 2017, 05:51 AM   #1
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Which f(x) to use?

I hope you can read the questions clearly:

For x = -2 and x = 0, which f(x) to use for each of them? I am confused because both of them are turning points.
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November 12th, 2017, 06:26 AM   #2
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I think we can't read it ... However, I can't.
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November 12th, 2017, 07:17 AM   #3
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$f(x)$ is continuous so you can just read off the graph. You don't need the slope.

Last edited by skipjack; November 12th, 2017 at 08:14 AM.
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November 12th, 2017, 09:12 AM   #4
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See a clearer image below.
Question13.jpg
(b)
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November 12th, 2017, 05:22 PM   #5
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Quote:
Originally Posted by v8archie View Post
$f(x)$ is continuous so you can just read off the graph. You don't need the slope.
What do you mean?
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November 12th, 2017, 06:04 PM   #6
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Just look at the graph. $f(-2)=0$ and $f(0)=2$.
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November 12th, 2017, 07:38 PM   #7
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So, (g o f o h)(-2) = g(0) and (g o f o h)(0) = g(2)?
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November 12th, 2017, 08:44 PM   #8
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As h(-2) = 5 and f(5) = 2, you should have obtained g(2) (which is 5), not g(0).
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November 12th, 2017, 08:54 PM   #9
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\begin{align*}&&(g \circ f \circ h)(-2) &= g\bigg(f\big(h(-2)\big)\bigg) \\
h(x)=-x+3 \implies h(-2) &= -(-2)+3 \\ &= 2+3 = 5 &\implies (g \circ f \circ h)(-2) &= g\big(f(5)\big) \\
f(5) &= 2 & \implies (g \circ f \circ h)(-2) &= g(2) & \text{(read from the graph)} \\
g(x) = 2x + 1 \implies g(2) &= 2(2)+1 \\ &= 5 &\implies (g \circ f \circ h)(-2) &= 5 \\
\end{align*}
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November 12th, 2017, 10:07 PM   #10
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I see, guess I made some mistakes last night.
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