Please help me with my Pre Calculus Homework!

Use the following functions for questions 1-7: f(x)=2x+3; g(x)=x^2-1; and h(x)=2x^2+3x-1
1) f(3)=
2) g(-4)=
3) (f•g)(2)=
4) (f+g)(3)=
5) h(g(2))=
6) h•g•f(-5/2)=
7) Write the composition function that would result in (2x+3)^2-1.

f(x)=2x+3

1.) f(3) = 2(3) + 3 = 6 + 3 = 9

2.) g(-4) = (-4)^2 - 1 = 16 - 1 = 15

3.) (f•g)(2)= f(2) * g(2) = [2(2) + 3] * [(2)^2 - 1] = 7 * 3 = 21

4.) (f+g)(3)= f(3) + g(3) = [ 2(3) + 3] + [(3)^2 - 1] = 9 + 8 = 17

5.) h(g(2))

Step 1: Find g(2)

g(2) = (2)^2 - 1 = 3

So

h(g(2)) = h(3)

h(3) = 2(3)^2 + 3(2) - 1

= 18 + 6 - 1

= 23

so h(g(2)) = h(3) = 23

6.) Works like question 4 but now you have 3 things to multiply [h(-5/2) * g(-5/2) * f(-5/2)]

7.) g(f(x)) = g(2x+3) = [(2x+3)^2] - 1

So it is g(f(x).

[You can compute f(g(x)) (if you want to) to see why that is not right]

tions for question 6: f(x)=2x+3; g(x)=x^2-1; and h(x)=2x^2+3x-1
6) h•g•f(-5/2)=

h * g* f (-5/2) = h(-5/2) * g(-5/2) * f(-5/2)

= [ 2* (-5/2) + 3] * [ (-5/2)^2 - 1] * 2(-5/2)^2 + 3(-5/2)

= -42

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sixty two divded by 8 is the sum of the difference of the intervals equated to the necessity of that function. furthermore, the reasoning of the legimate factors in the denomenator would negate it such that the sin and cosine of that particular function become lost.