[tex]A)\\f(x)=5x+1\\\\f(7):\ \text{Put x = 7 to the equation of the function:}\\\\f(7)=5(7)+1=35+1=36\\\\\boxed{f(7)=36}[/tex]
[tex]B)\\f(x)=5x+1\to y=5x+1\\\\\text{exchange x to y}\\\\x=5y+1\\\\\text{solve for y}\\\\5y+1=x\qquad\text{subtract 1 from both sides}\\\\5y=x-1\qquad\text{divide both sides by 5}\\\\y=\dfrac{1}{5}x-\dfrac{1}{5}\\\\\boxed{f^{-1}(x)=\dfrac{1}{5}x-\dfrac{1}{5}}[/tex]
[tex]C)\\f^{-1}(x)=\dfrac{1}{5}x-\dfrac{1}{5}\\\\f^{-1}(7):\ \text{Put x = 7 to the equation of the function}\ f^{-1}(x):\\\\f^{-1}(7)=\dfrac{1}{5}(7)-\dfrac{1}{5}=\dfrac{7}{5}-\dfrac{1}{5}=\dfrac{7-1}{5}=\dfrac{6}{5}=1\dfrac{1}{5}\\\\\boxed{f^{-1}(7)=1\dfrac{1}{5}}[/tex]
[tex]D)\\f(f^{-1}(7))\\\\f^{-1}(7)=\dfrac{6}{5},\ \text{Therefore put}\ x=\dfrac{6}{5}\ \text{to the equation of the function}\ f(x):\\\\f\left(\dfrac{6}{5}\right)=5\left(\dfrac{6}{5}\right)+1=6+1=7\\\\\boxed{f(f^{-1}(7))=7}[/tex]
Answer:
A . 36
B. [tex]f^{-1}(x)=\frac{1}{5}(x-1)[/tex]
C. 6/5
D. 7
Step-by-step explanation:
We have been given the function [tex]f(x)=5x+1[/tex]
A. Substitute x = 7 in the given function
[tex]f(7)=5(7)+1\\f(7)=36[/tex]
B. Let us find the inverse of the function
[tex]f(x)=5x+1\\y=5x+1[/tex]
Interchange x and y, we get
[tex]x=5y+1[/tex]
Solve for y
[tex]x-1=5y\\\\y=f^{-1}(x)=\frac{1}{5}(x-1)[/tex]
C. Substitute x= 7, in the inverse function
[tex]f^{-1}(7)=\frac{1}{5}(7-1)\\\\f^{-1}(7)=\frac{6}{5}[/tex]
D. Substitute x = 6/5 in the given function, we get
[tex]f(f^{-1}(7))=5\cdot \frac{6}{5} +1\\\\f(f^{-1}(7))=6+1\\\\f(f^{-1}(7))=7[/tex]
