2Alecture2 - Math 2A Lecture 2 Section 1.3 New functions...

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Math 2A: Lecture 2 Section 1.3: New functions from old functions
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Quiz : The black graph is y=sin(x). The red graph is: y=sin(x/2) y=sin(2x) y=sin(x-1)
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The black graph is y=sin(x). The red graph is: y=sin(2x) Answer :
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Quiz : To compute f o g you apply: first f, then g first g, then f
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Answer : To compute f o g you apply first g, then f
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Composition of functions f g f g x g(x) f(g(x)) To find f g you should: take the formula for f(x), and replace every x by g(x) f(x)=cos(x) g(x)=3x 2 +1 f o g(x)=f(g(x))=cos(3x 2 +1)
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f(x) = x 2 +1 g(x) = tan x f(g(x)) = (tan x ) 2 +1 g(f(x)) = tan(x 2 +1) g composed with f (evaluated at x) f composed with g (evaluated at x)
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Quiz : Given F(x)= (2+1/x) 3, find f and g such that F=f o g f(x)=(2+x) 3 g(x)=1/x f(x)=2+1/x g(x)=x 3
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Answer : Given F(x)= (2+1/x) 3 , we can write F=f o g with f(x)=(2+x) 3 g(x)=1/x Note that f is the outer function and g is the inner function . If you choose f(x)=2+1/x g(x)=x 3 , then you get f o g(x)= =2+1/x 3 (which is not F(x)!)
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graph of y=f(x+c) Translating graphs graph of y=f(x) graph of y=f(x-c) move c units to the right move c units to the left graph of y=f(x)-c graph of y=f(x)+c move c units upward move c units downward
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y=f(x+c) y=f(x) y=f(x-c) right left y=f(x)-c y=f(x)+c upward downward
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y=f(x+c) y=f(x) y=f(x-c) right left y=f(x)-c y=f(x)+c upward downward
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The equation of the red curve is… y=(x+2) 2 + 3 y=(x-2) 2 + 3 I am not sure Quiz:
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Answer: The equation of the red curve is y=(x-2) 2 + 3
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