Question

For f(x) = 5x + 1

Answer 1

Answer:

A. $f(7) = 5(7) + 1\\\\f(7) = 36$

B. $f^{-1}(x) = \frac{x - 1}{5}$

C. $f^{-1}(7) =\frac{6}{5}$

D. $f(\frac{6}{5}) = 7$

Step-by-step explanation:

A. To solve the first part of the problem we must replace $x = 7$ in the function $f(x) = 5x + 1$

So:

$f(7) = 5(7) + 1\\\\f(7) = 36$

B. In part B we must find the inverse function of $f(x) = 5x + 1$

To find the inverse function do $y = f(x)$

$y = 5x +1$

Now clear the variable x.

$\frac{y - 1}{5} = x$

Replace x with y.

$y = \frac{x - 1}{5}$

Finally

$f^{-1}(x) = \frac{x - 1}{5}$

C. Now we take the inverse function found above and replace $x = 7$

$f^{-1}(7) = \frac{7 - 1}{5}\\\\f(7) = \frac{6}{5}$

D. Now we substitute $x = f(f^{-1}(7))$ in the original function.

$x = f( f^{-1}(7))\\\\f^{-1}(7) = \frac{6}{5}\\\\ x= f(\frac{6}{5} )\\\\f(\frac{6}{5}) = 5(\frac{6}{5}) + 1\\\\f(\frac{6}{5}) = 7$

Answer 2

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)):

$f(x) = 5x + 1$

A. Find f(7):

$f(7)=5(7)+1 = 36$

B. Find f'1(x):

$y = 5x + 1$

Making x the subject to get:

$x = \frac{y-1}{5}$

$f'1(x) = \frac{x-1}{5}$

C. Find f−1(7):

$f'1(7) = \frac {7-1} {5} = \frac {6} {5}$

D. Find f( f−1(7)):

$x = f ( f^{-1} (7)) \\\\ f^{-1} (7) = \frac{6} {5} \\\\ x = f (\frac {6} {5} ) \\\\ f(\frac {6} {5}) = 5 (\frac {6} {5}) + 1 \\\\ f ( \frac {6} {5}) = 7$