Question

Pre calculus 4b. The function g is defined as g(x) = InX, X E R^tThe graph of y = g(x) and the graph of y = f^-1 (x) intersect at the point P.Find the exact coordinates of P.

Answer 1

Part A.

Wrtite out the function given

`f(x)=e^(3x+1)`

A functionn g(x) is the inverse of f(x) if for y=f(x), x=g(y).

To obtain the inverse of f(x), replace f(x)=y, we have

`\begin{gathered} f(x)=e^(3x+1) \\ y=e^(3x+1) \end{gathered}`

Isolate x in the equation above

take logarithm of both sides

`\begin{gathered} \text{lny}=\ln e^(3x+1) \\ \text{ Since ln e=1} \\ \text{Then} \\ \ln y=3x+1 \end{gathered}`

Subrtract 1 from both sides, we have

`\begin{gathered} \ln y-1=3x+1-1 \\ \text{Then} \\ \ln y-1=3x \\ \text{Divide both sides by 3, we have } \\ (\ln y-1)/(3)=(3x)/(3) \end{gathered}`

Then

`x=(\ln y-1)/(3)`

Then the inverse of the function becomes

`f^(-1)(x)=(\ln x-1)/(3)`

Hence

The inverse of the function, f(x) is

f¹(x)=ln x - 1 / 3

The Domian of a function are the set of the input value of for which the function is defined or real.

Hence

`\begin{gathered} \text{The domain is } \\ (0,\infty) \end{gathered}`

Therefore

The domain fo the inverse of f(x) is (0, ∞)

Part B

The function g(x) is given as

`g(x)=\ln x`
`f^(-1)(x)=(\ln x-1)/(3)`

The point of intersection is the point where

`f^(-1)(x)=g(x)`

Then, we have

`\begin{gathered} lnx=\: (\ln\:x-1)/(3) \\ \text{Multiply both sides by 3} \\ 3\text{ lnx=lnx-1} \end{gathered}`

Then, subtract both sides by ln x

`\begin{gathered} 3\ln x-\ln x=\ln x-\ln x-1 \\ 2\ln x=-1 \end{gathered}`

Divide both sides with 2 we have

`\begin{gathered} (2\ln x)/(2)=-(1)/(2) \\ \text{Then} \\ \ln x=-(1)/(2) \end{gathered}`

Then, take the exponent of both sides, we have

`\begin{gathered} e^(\ln x)=e^{-(1)/(2)} \\ x=e^{-(1)/(2)} \end{gathered}`

Then

`\begin{gathered} x=\frac{1}{e^{(1)/(2)}}=\frac{1}{\sqrt[]{e}} \\ \text{Hence } \\ x=\frac{1}{\sqrt[]{e}} \end{gathered}`

Then substitute the value of x into g(x) to find y, we have \

From the question,

`\begin{gathered} y=g(x) \\ \text{and } \\ g(x)=\ln x \\ \text{Then} \\ y=\ln x \\ \text{ since x=e}^{-(1)/(2)} \\ \text{Then} \\ y=\ln e^{-(1)/(2)}=-(1)/(2) \end{gathered}`

Therefore the point of intersection P is

`(\frac{1}{\sqrt[]{e}},-(1)/(2))`

Consider the graph below

Therefore point P is (0.607, -0.5)