Question

Part 1. Using the two functions listed below, insert numbers in place of the letters a, b, c, and d so that f(x) and g(x) are inverses.

f(x)=
x+a
b

g(x)=cx−d

Part 2. Show your work to prove that the inverse of f(x) is g(x).

Part 3. Show your work to evaluate g(f(x)).

Part 4. Graph your two functions on a coordinate plane. Include a table of values for each function. Include 5 values for each function. Graph the line y = x on the same graph.

Answer 1

what is f(x)?

x + ab?

Answer 2

or possibly x + a
-----
b

Answer 3

hint:
if g is the inverse function of f, then we can write:
\[\huge \begin{gathered}
g\left( {f\left( x \right)} \right) = x \hfill \\
\hfill \\
f\left( {g\left( y \right)} \right) = y \hfill \\
\end{gathered} \]

Answer 4

can someone teach me how to solve it step by step? coz i have no idea what to do

Answer 5

we can do that but we need to know exactly what the question is.

Answer 6

oh what do u mean is the question confusing

Answer 7

yes it looks to me that the first function is probably
f(x) = x + a
-----
b

but i wanted to confirm it.

Answer 8

from the first condition:
\[\Large \Large g\left( {f\left( x \right)} \right) = x\]
by substitution, I get:
\[\Large g\left( {f\left( x \right)} \right) = c\left( {f\left( x \right)} \right) - d = c\left( {x + a} \right) - d = cx + ca - d\]
so we have to solve this equation:
\[\Large cx + ca - d = x\]

Answer 9

yes i guess when i copyd it down the function didn't come out right my bad sorry

Answer 10

rthats ok
sorry but i have to go right now

Answer 11

please solve that equation, using the principle of identity of polynomials

Answer 12

thats ok

Answer 13

oops.. I have neglected the coefficient b. Here are the right equations:

\[\large \begin{gathered}
g\left( {f\left( x \right)} \right) = c\left( {f\left( x \right)} \right) - d = c\left( {\frac{{x + a}}{b}} \right) - d = \frac{{cx}}{b} + \frac{{ca}}{b} - d \hfill \\
\frac{{cx}}{b} + \frac{{ca}}{b} - d = x \hfill \\
\end{gathered} \]

Answer 14

is it \[x= - \left[\begin{matrix}ac-d & ? \\ c-1? & ?\end{matrix}\right]\]

Answer 15

hint:
if we apply the principle of identity of polynomials, we can write:
\[\Large \frac{c}{b} = 1,\quad \frac{{ca}}{b} - d = 0\]

Answer 16

i'm confused,sorry

Answer 17

we have to solve this eqution:
\[\Large \frac{{cx}}{b} + \frac{{ca}}{b} - d = x\]

Answer 18

now, I apply the principle of identity of polynomials, and I say that the polynomial at the left side is identical to the polynomial at the right side, if the corresponding coefficients are equal, namely
\[\Large \begin{gathered}
\frac{c}{b} = 1,\quad \left( {{\text{for the coefficients of }}x} \right) \hfill \\
\hfill \\
\frac{{ca}}{b} - d = 0\quad \left( {{\text{for the known terms}}} \right) \hfill \\
\end{gathered} \]

Answer 19

oh ok

Answer 20

here is another way to write the equation above:
\[\Large \frac{{cx}}{b} + \frac{{ca}}{b} - d = 1 \cdot x + 0\]

Answer 21

so, from the conditions above, I get:
\[\Large c = b,\quad a = d\]