Answer:
yes it does
Step-by-step explanation:
Answer:
small pizza
Step-by-step explanation:
1/2 gabby 1/2 sam
Answer:
![\huge\boxed{a=9 ; b = -8}](https://tex.z-dn.net/?f=%5Chuge%5Cboxed%7Ba%3D9%20%3B%20b%20%3D%20-8%7D)
Step-by-step explanation:
![f(x) = \frac{ax+b}{x}](https://tex.z-dn.net/?f=f%28x%29%20%3D%20%5Cfrac%7Bax%2Bb%7D%7Bx%7D)
Putting x = 1
=> ![f(1) = \frac{a(1)+b}{1}](https://tex.z-dn.net/?f=f%281%29%20%3D%20%5Cfrac%7Ba%281%29%2Bb%7D%7B1%7D)
Given that f(1) = 1
=> ![1 = a + b](https://tex.z-dn.net/?f=1%20%3D%20a%20%2B%20b)
=>
-------------------(1)
Now,
Putting x = 2
=> ![f(2) = \frac{a(2)+b}{2}](https://tex.z-dn.net/?f=f%282%29%20%3D%20%5Cfrac%7Ba%282%29%2Bb%7D%7B2%7D)
Given that f(2) = 5
=> ![5 = \frac{2a+b}{2}](https://tex.z-dn.net/?f=5%20%3D%20%5Cfrac%7B2a%2Bb%7D%7B2%7D)
=> ![2a+b = 5*2](https://tex.z-dn.net/?f=2a%2Bb%20%3D%205%2A2)
=>
----------------(2)
Subtracting (2) from (1)
![a+b-(2a+b) = 1-10\\a+b-2a-b = -9\\a-2a = -9\\-a = -9\\a = 9](https://tex.z-dn.net/?f=a%2Bb-%282a%2Bb%29%20%3D%201-10%5C%5Ca%2Bb-2a-b%20%3D%20-9%5C%5Ca-2a%20%3D%20-9%5C%5C-a%20%3D%20-9%5C%5Ca%20%3D%209)
For b , Put a = 9 in equation (1)
![9+b = 1\\Subtracting \ both \ sides \ by \ 9\\b = 1-9\\b = -8](https://tex.z-dn.net/?f=9%2Bb%20%3D%201%5C%5CSubtracting%20%5C%20both%20%5C%20sides%20%5C%20by%20%5C%209%5C%5Cb%20%3D%201-9%5C%5Cb%20%3D%20-8)
Answer:
![(f\ o\ g)(x) = 3x](https://tex.z-dn.net/?f=%28f%5C%20o%5C%20g%29%28x%29%20%3D%203x)
![-\infty < x < \infty](https://tex.z-dn.net/?f=-%5Cinfty%20%3C%20x%20%3C%20%5Cinfty)
Step-by-step explanation:
Given
![f(x) = ln(x^2)](https://tex.z-dn.net/?f=f%28x%29%20%3D%20ln%28x%5E2%29)
![g(x)=\sqrt{e^{3x}}](https://tex.z-dn.net/?f=g%28x%29%3D%5Csqrt%7Be%5E%7B3x%7D%7D)
Solving (a): (f o g)(x)
This is calculated as:
![(f\ o\ g)(x) = f(g(x))](https://tex.z-dn.net/?f=%28f%5C%20o%5C%20g%29%28x%29%20%3D%20f%28g%28x%29%29)
We have:
![f(x) = ln(x^2)](https://tex.z-dn.net/?f=f%28x%29%20%3D%20ln%28x%5E2%29)
![f(g(x)) = \ln((g(x))^2)](https://tex.z-dn.net/?f=f%28g%28x%29%29%20%3D%20%20%5Cln%28%28g%28x%29%29%5E2%29)
Substitute: ![g(x)=\sqrt{e^{3x}}](https://tex.z-dn.net/?f=g%28x%29%3D%5Csqrt%7Be%5E%7B3x%7D%7D)
![f(g(x)) = \ln(\sqrt{e^{3x}})^2](https://tex.z-dn.net/?f=f%28g%28x%29%29%20%3D%20%20%5Cln%28%5Csqrt%7Be%5E%7B3x%7D%7D%29%5E2)
Evaluate the square
![f(g(x)) = \ln(e^{3x})](https://tex.z-dn.net/?f=f%28g%28x%29%29%20%3D%20%20%5Cln%28e%5E%7B3x%7D%29)
Using laws of natural logarithm:
![\ln(e^{ax}) = ax](https://tex.z-dn.net/?f=%5Cln%28e%5E%7Bax%7D%29%20%3D%20ax)
So:
![f(g(x)) = \ln(e^{3x})](https://tex.z-dn.net/?f=f%28g%28x%29%29%20%3D%20%20%5Cln%28e%5E%7B3x%7D%29)
![f(g(x)) = 3x](https://tex.z-dn.net/?f=f%28g%28x%29%29%20%3D%20%203x)
Hence:
![(f\ o\ g)(x) = 3x](https://tex.z-dn.net/?f=%28f%5C%20o%5C%20g%29%28x%29%20%3D%203x)
Solving (b): The domain
We have:
![f(g(x)) = 3x](https://tex.z-dn.net/?f=f%28g%28x%29%29%20%3D%20%203x)
The above function has does not have any undefined points and domain constraints.
<em>Hence, the domain is: </em>