Answer:
C and D
Step-by-step explanation:
We have to find pairs of functions which are inverses of each other.
1.[tex]f(x)=2x+3[/tex] and [tex]g(x)=0.5x-3[/tex]
Suppose y=f(x)
[tex]y=2x+3[/tex]
[tex]2x=y-3[/tex]
[tex]x=\frac{y-3}{2}[/tex]
[tex]x=f^{-1}(y)[/tex]
[tex]f^{-1}(y)=\frac{y-3}{2}[/tex]
y replace by x
Then , we get
[tex]f^{-1}(x)=\frac[x-3}{2}[/tex]
Hence, f(x) and g(x) are not inverses to each other.
B.[tex]f(x)=\frac{1}{6}x-2[/tex], and [tex]g(x)=6x-12[/tex]
Suppose,[tex]y=f(x)=\frac{1}{6}x-2[/tex]
[tex]y+2=\frac{1}{6}x[/tex]
[tex]6y+12=x[/tex]
Therefore,[tex]f^{-1}(x)=6x+12[/tex]
Hence, f(x) and g(x) are not inverses to each other.
C.[tex]f(x)=\frac{1}{3}x+5[/tex] and [tex]g(x)=3x-15[/tex]
Suppose ,[tex]y=f(x)=\frac{1}{3}x+5[/tex]
[tex]y-5=\frac{1}{3}x[/tex]
[tex]3y-15=x[/tex]
Therefore, [tex]f^{-1}(x)=3x-15=g(x)[/tex]
Hence, f(x) and g(x) are inverses of each other.
D.[tex]f(x)=x^2+7[/tex] and [tex]g(x)=\pm \sqrt{x-7}[/tex]
Suppose , [tex]y=f(x)=x^2+7[/tex]
[tex]y-7=x^2[/tex]
[tex]x=\pm \sqrt{x-7}[/tex]
Therefore, [tex]f^{-1}(x)=\pm\sqrt{x-7}=g(x)[/tex]
Hence, f(x) and g(x) are inverses to each other.
