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 November 12th, 2017, 05:51 AM #1 Senior Member     Joined: Nov 2010 From: Indonesia Posts: 1,672 Thanks: 120 Math Focus: Trigonometry and Logarithm 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.
 November 12th, 2017, 06:26 AM #2 Member   Joined: Oct 2016 From: Slovenia, Europe Posts: 52 Thanks: 5 I think we can't read it ... However, I can't.
 November 12th, 2017, 07:17 AM #3 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,119 Thanks: 2369 Math Focus: Mainly analysis and algebra $f(x)$ 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.
 November 12th, 2017, 09:12 AM #4 Global Moderator   Joined: Dec 2006 Posts: 18,445 Thanks: 1462 See a clearer image below. Question13.jpg (b) k.jpg Thanks from topsquark
November 12th, 2017, 05:22 PM   #5
Senior Member

Joined: Nov 2010
From: Indonesia

Posts: 1,672
Thanks: 120

Math Focus: Trigonometry and Logarithm
Quote:
 Originally Posted by v8archie $f(x)$ is continuous so you can just read off the graph. You don't need the slope.
What do you mean?

 November 12th, 2017, 06:04 PM #6 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,119 Thanks: 2369 Math Focus: Mainly analysis and algebra Just look at the graph. $f(-2)=0$ and $f(0)=2$.
 November 12th, 2017, 07:38 PM #7 Senior Member     Joined: Nov 2010 From: Indonesia Posts: 1,672 Thanks: 120 Math Focus: Trigonometry and Logarithm So, (g o f o h)(-2) = g(0) and (g o f o h)(0) = g(2)?
 November 12th, 2017, 08:44 PM #8 Global Moderator   Joined: Dec 2006 Posts: 18,445 Thanks: 1462 As h(-2) = 5 and f(5) = 2, you should have obtained g(2) (which is 5), not g(0).
 November 12th, 2017, 08:54 PM #9 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,119 Thanks: 2369 Math Focus: Mainly analysis and algebra \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*}
 November 12th, 2017, 10:07 PM #10 Senior Member     Joined: Nov 2010 From: Indonesia Posts: 1,672 Thanks: 120 Math Focus: Trigonometry and Logarithm I see, guess I made some mistakes last night.

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