Question 1201983
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The rule of this website is to post <b><u>one</u></b> problem at a time.


I'll do questions (1) through (4) to get you started.


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Problem 1


f(x) = 2x
f(3) = 2*3
f(3) = 6
and
g(x) = x-5
g(3) = 3-5
g(3) = -2
Therefore,
(f+g)(3) = f(3)+g(3) = 6 + (-2) = 4


Here's another approach:
(f+g)(x) = f(x)+g(x)
(f+g)(x) = ( 2x ) + ( x-5 )
(f+g)(x) = 3x-5
(f+g)(3) = 3*3-5
(f+g)(3) = 9-5
(f+g)(3) = 4



Answer: <font color=red size=4>4</font>


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Problem 2


Method 1
f(x) = 2x
f(-1) = 2*(-1)
f(-1) = -2
and
h(x) = x^2-1
h(-1) = (-1)^2-1
h(-1) = 1-1
h(-1) = 0
Both of those lead to
(f-h)(-1) = f(-1) - h(-1)
(f-h)(-1) = -2 - 0
(f-h)(-1) = -2



Method 2
(f-h)(x) = f(x)-h(x)
(f-h)(x) = (2x) - (x^2-1)
(f-h)(x) = 2x - x^2+1
(f-h)(x) = -x^2 + 2x + 1
(f-h)(-1) = -(-1)^2 + 2(-1) + 1
(f-h)(-1) = -(1) + 2(-1) + 1
(f-h)(-1) = -1 - 2 + 1
(f-h)(-1) = -3 + 1
(f-h)(-1) = -2



Answer: <font color=red size=4>-2</font>


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Problem 3


(h-g)(x) = h(x) - g(x)
(h-g)(x) = [ h(x) ] - [ g(x) ]
(h-g)(x) = [ x^2-1 ] - [ x-5 ]
(h-g)(x) =  x^2-1  -  x+5 
(h-g)(x) =  x^2 - x + 4 


Answer: <font color=red size=4>x^2 - x + 4</font>


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Problem 4


(fg)(x) = f(x)*g(x)
(fg)(x) = [ f(x) ] * [ g(x) ]
(fg)(x) = [ 2x ] * [ x-5 ]
(fg)(x) = 2x(x-5)
(fg)(x) = 2x*x + 2x*(-5)
(fg)(x) = 2x^2 - 10x


Answer: <font color=red size=4>2x^2 - 10x</font>
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