Question

Which of the following points lies on the graph of the inverse of f(x) = 6^x+1 (-1,1) (-1,0)(1,-1)

Answer 1

We will investigate the relevance of inverse of a function.

A function is usually defined as a relationhsip between the input variable and output. It relates the two variables in form off an elgebraic expression. For every value of input ( x ) their is a unique value of the output this qualifies as a definition of a function.

We are given the following function as follows:

`\textcolor{#FF7968}{f(x)=6^{x\text{ +1}}}\text{\textcolor{#FF7968}{ }}`

Where,

Next we will go ahead an define an inverse of a function. The term inverse is given to a transformation of any line or a graph undergoing reflection. Strictly, speaking in terms of a function an inverse is:

`Inverse\colon\text{Reflection across ( y = x ) line on a graph}`

The above expression says that an inverse of any function is a reflection of the entire graph across the ( y = x ) line on a 2D cartesian coordinate system.

Mathematically evaluating the reflection across any line we will follow the guidelines:

Step 1: Interchange of input and out variables

This step basically means that when we look at our given function f ( x ) we have two variables i.e Independent variable ( x ) and dependent variable ( f (x) ). We will interchange these with each other using the following relation of reflection:

`\begin{gathered} \text{Reflection across ( y = x ) line:} \\ \textcolor{#FF7968}{x\to f^(-1)}\text{\textcolor{#FF7968}{ ( x ) AND f ( x ) }}\textcolor{#FF7968}{\to}\text{\textcolor{#FF7968}{ x}} \end{gathered}`

We will use the above two relations highlighted in red and do the following:

`\textcolor{#FF7968}{x=6^{f^(-1)(x)\text{ + 1}}}`

Step 2: Make the inverse notation subject of the equation

Once we have interchanged the vairables we will make f^-1 ( x ) the subject of an equation by first taking natural logs on both sides ( Ln ) as follows:

`Ln(x)=Ln(6)^{f^(-1)(x)\text{ + 1}}`

Next we will use the natural log base law as follows:

`Ln(b)^c\text{ = c}\cdot Ln\text{ ( b ) }\ldots\text{ Log power to base law}`

Apply the above law to the right hand side of the latest expression formed above:

`\textcolor{#FF7968}{Ln(x)=(f^(-1)(x)}\text{\textcolor{#FF7968}{ + 1 )}}\textcolor{#FF7968}{\cdot Ln(6)}`

Now we will divide the Ln ( 6 ) on both sides of the equation as follows:

`\begin{gathered} \frac{Ln(x)}{Ln\text{ ( 6 )}}=(f^(-1)(x)\text{ + 1 )}\cdot(Ln(6))/(Ln(6)) \\ \\ \textcolor{#FF7968}{\frac{Ln(x)}{Ln\text{ ( 6 )}}=f^(-1)(x)}\text{\textcolor{#FF7968}{ + 1 }} \end{gathered}`

We will now subtract ( 1 ) on both sides as follows:

`\begin{gathered} (Ln(x))/(Ln(6))-1=f^(-1)(x)\text{ + 1 - 1} \\ \\ \textcolor{#FF7968}{(Ln(x))/(Ln(6))-1=f^(-1)(x)} \end{gathered}`

The above expression is the final expression for an inverse of a function. The inverse of the given function is:

`\textcolor{#FF7968}{f^(-1)(x)=(Ln(x))/(Ln(6))-1}`

Now we will investigate what point will lie on the inverse of the function graph. Before we go to that we will recall some properties of a Natural Log as follows:

`\textcolor{#FF7968}{Ln}\text{\textcolor{#FF7968}{ ( x ), domain: x}}\textcolor{#FF7968}{\ge}\text{\textcolor{#FF7968}{1}}`

The above is the most important property of a natural log. This means for any values of ( x ) less than ( 1 ) the function is undefined or there is no such point!

Using the above property we will rule out most of our option choices as follows:

`\begin{gathered} \text{\textcolor{#FF7968}{Option A: ( -1 , 1 ) }}\textcolor{#FF7968}{\ldots}\text{\textcolor{#FF7968}{ Rejected because x < 1}} \\ \text{\textcolor{#FF7968}{Option B: (-1,0) }}\textcolor{#FF7968}{\ldots}\text{\textcolor{#FF7968}{ Reected because x < 1}} \end{gathered}`

Now we will check about option ( C ) whether the point satisfies our inverse relationship or not as follows:

`\begin{gathered} \text{Option C: ( 1 , -1 ) }\ldots\text{ check!} \\ \\ f^(-1)(\text{ x ) = }(Ln(x))/(Ln(6))\text{ - 1 }\ldots\text{ Substitute both x and y coordinates} \\ \\ -1\text{ = }(Ln(1))/(Ln(6))\text{ - 1 } \\ \\ -1\text{ = 0 - 1} \\ \textcolor{#FF7968}{-1}\text{\textcolor{#FF7968}{ = -1 }}\textcolor{#FF7968}{\ldots}\text{\textcolor{#FF7968}{ Consistency check! Hence, option is correct!}} \end{gathered}`

Therefore, the last option ( C ) lies on the inverse of the given function:

`\text{\textcolor{#FF7968}{Option C: ( 1 , -1 )}}`