Answer:
D is the answer,that is the incorrect inverse function.
Step-by-step explanation:
For the first one
g(x)=6x-1
Let y=6x-1
6x=y+1
x=y+1/6
Therefore the inverse equation will be
x+1/6 so,the first equation is right.
The second one
g(x)=19x+4
Let y=19x+4
19x=y-4
x=y-4/19 so,the second function is right.
For the last one
f(x)=x/x+20
y=x/x+20
y(x+20)=x
x+20=x/y
x=x/y-20
g(x)=x/y-20,so the function is wrong and that is the aswer.
Answer:
The pair of functions that is not inverse functions are:
[tex]f(x)=\dfrac{x}{x+20}\ and\ g(x)=\dfrac{20x}{x-1}[/tex]
Step-by-step explanation:
We know that two functions a and b are said to be inverse of each other if:
f(g(x))=g(f(x))=x
i.e. the compositions of the function gives identity.
and the inverse of the given function is calculated by equation the function to y and then calculating the value of x in terms of y.
and the function in terms of y is the inverse function.
A)
[tex]f(x)=\dfrac{x+1}{6}[/tex]
Now if f(x)=y then
[tex]\dfrac{x+1}{6}=y\\\\\\x+1=6y\\\\\\x=6y-1\\\\\\g(x)=6x-1[/tex]
B)
[tex]f(x)=\dfrac{x-4}{19}\\\\\\f(x)=y\\\\\\\dfrac{x-4}{19}=y\\\\\\x=19y+4[/tex]
i.e.
[tex]g(x)=19x+4[/tex]
C)
[tex]f(x)=x^5\\\\\\f(x)=y\\\\\\x^5=y\\\\\\x=\sqrt[5]{y}[/tex]
i.e. we have:
[tex]g(x)=\sqrt[5]{x}[/tex]
D)
[tex]f(x)=\dfrac{x}{x+20}[/tex]
if f(x)=y then
[tex]\dfrac{x}{x+20}=y\\\\\\x=xy+20y\\\\\\x(1-y)=20y\\\\x=\dfrac{20y}{1-y}[/tex]
[tex]g(x)=\dfrac{20x}{1-x}\neq \dfrac{20x}{x-1}[/tex]
Hence, option: D is the answer.
