Problem 1
<h3>Answer: False</h3>
---------------------------------
Explanation:
The notation (f o g)(x) means f( g(x) ). Here g(x) is the inner function.
So,
f(x) = x+1
f( g(x) ) = g(x) + 1 .... replace every x with g(x)
f( g(x) ) = 6x+1 ... plug in g(x) = 6x
(f o g)(x) = 6x+1
Now let's flip things around
g(x) = 6x
g( f(x) ) = 6*( f(x) ) .... replace every x with f(x)
g( f(x) ) = 6(x+1) .... plug in f(x) = x+1
g( f(x) ) = 6x+6
(g o f)(x) = 6x+6
This shows that (f o g)(x) = (g o f)(x) is a false equation for the given f(x) and g(x) functions.
===============================================
Problem 2
<h3>Answer: True</h3>
---------------------------------
Explanation:
Let's say that g(x) produced a number that wasn't in the domain of f(x). This would mean that f( g(x) ) would be undefined.
For example, let
f(x) = 1/(x+2)
g(x) = -2
The g(x) function will always produce the output -2 regardless of what the input x is. Feeding that -2 output into f(x) leads to 1/(x+2) = 1/(-2+2) = 1/0 which is undefined.
So it's important that the outputs of g(x) line up with the domain of f(x). Outputs of g(x) must be valid inputs of f(x).
Answer:
g-1(x) = 3(x + 7).
Domain is All Real values of x.
Range is All Real values of the function.
Step-by-step explanation:
Let y = 1/3 x - 7
1/3 x = y + 7
Multiply both sides by 3:
x = 3(y + 7)
So the inverse of g(x) = g-1(x) = 3(x + 7).
Domain is All Real values of x.
Range is All Real values of the function.
Glad you know how to use the ^ feature :) it helps alot!
Answer: <span><span>4<span>(<span>x<span>−6</span></span>)</span></span><span>(<span>x+5</span>) </span></span>
3+2x
i think that would be the answer smol child
