The answer is d because it ends in an even number, and any number that ends with an even number is divisible by 2
<em><u>Method One</u></em>
f(g(x)) = x
<em><u>Method Two</u></em>
g(f(x)) = x
So let's pick a pair of functions and try this out.
f(x) = x^2 + 1
g(x) =sqrt(x - 1)
<em><u>Using Method 1</u></em>
f(g(x)) = (g(x)^2 + 1 You put a g(x) wherever you see an x in f(x)
f(g(x)) = [sqrt(x - 1)}^2 + 1 Substitute the right side of g(x) on the right side of f(x)
f(g(x)) = x - 1 + 1 Expand and cancel
f(g(x) = x
<em><u>Using Method 2</u></em>
g(f(x)) = sqrt(f(x) - 1) Put an f(x) wherever you see an x in g(x)
g(f(x)) = sqrt(x^2 + 1 - 1) Substitute the value of f(x) in the g(x) equation
g(f(x)) = sqrt(x^2) The 1s cancel. Take the square root of x^2
g(f(x)) = x You get x which is what you need to get.
So these two functions are the inverses of each other. Both methods confirm the results. A graph may help you to understand.
Notice how the red line (f(x) = x^2 + 1) is reflected across the green line to become the blue line (g(x) = sqrt(x - 1) ) That is another way to tell that 2 equations are inverses.
Note further that I have take the equations so that x in all three cases is ≥ 0
Answer:
---- ![x =\{-7,- 1\}](https://tex.z-dn.net/?f=x%20%3D%5C%7B-7%2C-%201%5C%7D)
---- ![x = \{1,7\}](https://tex.z-dn.net/?f=x%20%3D%20%5C%7B1%2C7%5C%7D)
----- ![x = \{-7,1\}](https://tex.z-dn.net/?f=x%20%3D%20%5C%7B-7%2C1%5C%7D)
---- ![x = \{-1,7\}](https://tex.z-dn.net/?f=x%20%3D%20%5C%7B-1%2C7%5C%7D)
Step-by-step explanation:
Given
The attached equations
Required
Match each with the solution
![(a)\ x + 4 = \±\sqrt9](https://tex.z-dn.net/?f=%28a%29%5C%20x%20%2B%204%20%3D%20%5C%C2%B1%5Csqrt9)
Take the square root of 9
![x + 4 = \±3](https://tex.z-dn.net/?f=x%20%2B%204%20%3D%20%5C%C2%B13)
Solve for x
![x =- 4 \±3](https://tex.z-dn.net/?f=x%20%3D-%204%20%20%5C%C2%B13)
Split
![x =\{- 4 -3,- 4 +3\}](https://tex.z-dn.net/?f=x%20%3D%5C%7B-%204%20-3%2C-%204%20%2B3%5C%7D)
![x =\{-7,- 1\}](https://tex.z-dn.net/?f=x%20%3D%5C%7B-7%2C-%201%5C%7D)
![(b)\ x - 4 = \±\sqrt9](https://tex.z-dn.net/?f=%28b%29%5C%20x%20-%204%20%3D%20%5C%C2%B1%5Csqrt9)
Take the square root of 9
![x - 4 = \±3](https://tex.z-dn.net/?f=x%20-%204%20%3D%20%5C%C2%B13)
Solve for x
![x = 4 \±3](https://tex.z-dn.net/?f=x%20%3D%204%20%5C%C2%B13)
Split
![x = \{4 -3,4+3\}](https://tex.z-dn.net/?f=x%20%3D%20%5C%7B4%20-3%2C4%2B3%5C%7D)
![x = \{1,7\}](https://tex.z-dn.net/?f=x%20%3D%20%5C%7B1%2C7%5C%7D)
![(c)\ x + 3 = \±\sqrt{16}](https://tex.z-dn.net/?f=%28c%29%5C%20x%20%2B%203%20%3D%20%5C%C2%B1%5Csqrt%7B16%7D)
Take the square root of 16
![x + 3 = \±4](https://tex.z-dn.net/?f=x%20%2B%203%20%3D%20%5C%C2%B14)
Solve for x
![x =- 3 \±4](https://tex.z-dn.net/?f=x%20%3D-%203%20%5C%C2%B14)
Split
![x = \{-3-4,-3+4\}](https://tex.z-dn.net/?f=x%20%3D%20%5C%7B-3-4%2C-3%2B4%5C%7D)
![x = \{-7,1\}](https://tex.z-dn.net/?f=x%20%3D%20%5C%7B-7%2C1%5C%7D)
![(d)\ x - 3 = \±\sqrt{16}](https://tex.z-dn.net/?f=%28d%29%5C%20x%20-%203%20%3D%20%5C%C2%B1%5Csqrt%7B16%7D)
Take the square root of 16
![x - 3 = \±4](https://tex.z-dn.net/?f=x%20-%203%20%3D%20%5C%C2%B14)
Solve for x
![x =3 \±4](https://tex.z-dn.net/?f=x%20%3D3%20%5C%C2%B14)
Split
![x = \{3-4,3+4\}](https://tex.z-dn.net/?f=x%20%3D%20%5C%7B3-4%2C3%2B4%5C%7D)
![x = \{-1,7\}](https://tex.z-dn.net/?f=x%20%3D%20%5C%7B-1%2C7%5C%7D)
The answer is D -1/2
As the equation relates to that answerr
So we're given Y = 3x.
We now can replace Y, for 3x in the equation:
Y = x + 4
3x = x + 4
3x - x = 4
2x = 4
x = 2