Answer:
A. [tex]f(7) = 5(7) + 1\\\\f(7) = 36[/tex]
B. [tex]f^{-1}(x) = \frac{x - 1}{5}[/tex]
C. [tex]f^{-1}(7) =\frac{6}{5}[/tex]
D. [tex]f(\frac{6}{5}) = 7[/tex]
Step-by-step explanation:
A. To solve the first part of the problem we must replace [tex]x = 7[/tex] in the function [tex]f(x) = 5x + 1[/tex]
So:
[tex]f(7) = 5(7) + 1\\\\f(7) = 36[/tex]
B. In part B we must find the inverse function of [tex]f(x) = 5x + 1[/tex]
To find the inverse function do [tex]y = f(x)[/tex]
[tex]y = 5x +1[/tex]
Now clear the variable x.
[tex]\frac{y - 1}{5} = x[/tex]
Replace x with y.
[tex]y = \frac{x - 1}{5}[/tex]
Finally
[tex]f^{-1}(x) = \frac{x - 1}{5}[/tex]
C. Now we take the inverse function found above and replace [tex]x = 7[/tex]
[tex]f^{-1}(7) = \frac{7 - 1}{5}\\\\f(7) = \frac{6}{5}[/tex]
D. Now we substitute [tex]x = f(f^{-1}(7))[/tex] in the original function.
[tex]x = f( f^{-1}(7))\\\\f^{-1}(7) = \frac{6}{5}\\\\ x= f(\frac{6}{5} )\\\\f(\frac{6}{5}) = 5(\frac{6}{5}) + 1\\\\f(\frac{6}{5}) = 7[/tex]
Answer with step-by-step explanation:
We are given the following function and we are to find f(7), f−1(x), f−1(7) and f( f−1(7)):
[tex] f(x) = 5x + 1 [/tex]
A. Find f(7):
[tex]f(7)=5(7)+1 = 36[/tex]
B. Find f'1(x):
[tex] y = 5x + 1 [/tex]
Making x the subject to get:
[tex] x = \frac{y-1}{5}[/tex]
[tex]f'1(x) = \frac{x-1}{5}[/tex]
C. Find f−1(7):
[tex] f'1(7) = \frac {7-1} {5} = \frac {6} {5} [/tex]
D. Find f( f−1(7)):
[tex] x = f ( f^{-1} (7)) \\\\ f^{-1} (7) = \frac{6} {5} \\\\ x = f (\frac {6} {5} ) \\\\ f(\frac {6} {5}) = 5 (\frac {6} {5}) + 1 \\\\ f ( \frac {6} {5}) = 7 [/tex]
