SOLUTION: Let f(x) = 2x, g(x) = x—5, and h(x)=x^2-1. Find each value. (1) (f+g)(3) (2) (f-h)(-1) (3) (h-g)(x) (4) (fg)(x) (5) (fh)(2) (6) (gh)(-2) (7) (f/h) (8) (h/g)(0

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Question 1201983: Let f(x) = 2x, g(x) = x—5, and h(x)=x^2-1. Find each value.
(1) (f+g)(3)
(2) (f-h)(-1)
(3) (h-g)(x)
(4) (fg)(x)
(5) (fh)(2)
(6) (gh)(-2)
(7) (f/h)
(8) (h/g)(0)
(9) (g/f)(x)
Answer by math_tutor2020(3816) (Show Source):
You can put this solution on YOUR website!

The rule of this website is to post one 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: 4

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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: -2

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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: x^2 - x + 4

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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: 2x^2 - 10x