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Chapter 4: Problem 5

Write an equation for the inverse of the function. $$ f(x)=e^{x-2} $$

Short Answer

Expert verified
f^{-1}(x) = \ln(x) + 2

Step by step solution

01

Write the function in terms of y

Start by rewriting the function with y instead of f(x). \[ y = e^{x-2} \]
02

Swap y and x

To find the inverse, interchange x and y.\[ x = e^{y-2} \]
03

Solve for y

Isolate y by taking the natural logarithm (ln) on both sides.\[ \ln(x) = y - 2 \]Add 2 to both sides:\[ y = \ln(x) + 2 \]
04

Write the inverse function

Replace y with the notation for the inverse function, which is f^{-1}(x).\[ f^{-1}(x) = \ln(x) + 2 \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Exponential Functions
Exponential functions are mathematical expressions in which a variable appears in the exponent. The general form is given as: \[ f(x) = a \times e^{bx} \] where:
  • **\(e\)** is the base of the natural logarithm, approximately equal to 2.71828.
  • **\(a\)** and **\(b\)** are constants.
  • **\(x\)** is the exponent.

For the function \( f(x)=e^{x-2} \), the operation is shifting the basic exponential function \( e^x \) horizontally by 2 units. By understanding the properties of exponential functions, you can deal with complex transformations and inverse functions more effectively.
Natural Logarithm
The natural logarithm, denoted as \( \ln(x) \), is the inverse function of the exponential function with base \(e\). This means that:\[ y = e^x \] is equivalent to:\[ x = \ln(y) \]
When you take the natural logarithm of both sides of an equation involving \(e\), it 'undoes' the exponentiation. For example, if you have an equation like \( x = e^{y-2} \), taking the natural logarithm of both sides leads to \( \ln(x) = y - 2 \). This property is crucial for solving for variables within exponential equations. In the exercise, taking the natural logarithm helped isolate \( y \), yielding \( y = \ln(x) + 2 \).
Function Transformation
Function transformations involve shifting, stretching, or reflecting the graph of a function. In the context of the given exercise, there are horizontal and vertical transformations involved. The given function is: \( f(x) = e^{x-2} \). The term \( x-2 \) indicates a horizontal shift to the right by 2 units from the basic exponential function \( e^x \). To find the inverse, transforming involves swapping x and y and then isolating y:
  • Starting with \( y = e^{x-2} \), we swap to get \( x = e^{y-2} \).
  • To isolate \( y \), we apply the natural logarithm: \( \ln(x) = y - 2 \).
  • Finally, we solve for \( y \) by adding 2 to both sides, giving us our inverse function: \( f^{-1}(x) = \ln(x) + 2 \).
Understanding these transformations is essential for mastering inverse functions, allowing you to switch perspectives between the function and its inverse easily.

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Most popular questions from this chapter

Solve the equation. Write the solution set with the exact solutions. Also give approximate solutions to 4 decimal places if necessary. \(\ln x+\ln (x-4)=\ln (3 x-10)\)

a. The populations of two countries are given for January 1,2000 , and for January 1,2010 . Write a function of the form \(P(t)=P_{0} e^{k t}\) to model each population \(P(t)\) (in millions) \(t\) years after January 1, 2000.$$ \begin{array}{|l|c|c|c|} \hline & \begin{array}{c} \text { Population } \\ \text { in 2000 } \\ \text { (millions) } \end{array} & \begin{array}{c} \text { Population } \\ \text { in 2010 } \\ \text { (millions) } \end{array} & \boldsymbol{P}(t)=\boldsymbol{P}_{0} e^{k t} \\ \hline \text { Switzerland } & 7.3 & 7.8 & \\ \hline \text { Israel } & 6.7 & 7.7 & \\ \hline \end{array}$$ b. Use the models from part (a) to predict the population on January \(1,2020,\) for each country. Round to the nearest hundred thousand. c. Israel had fewer people than Switzerland in the year 2000 , yet from the result of part (b), Israel will have more people in the year \(2020 ?\) Why? d. Use the models from part (a) to predict the year during which each population will reach 10 million if this trend continues.

Fluorodeoxyglucose is a derivative of glucose that contains the radionuclide fluorine- \(18\left({ }^{18} \mathrm{~F}\right) .\) A patient is given a sample of this material containing \(300 \mathrm{MBq}\) of \({ }^{18} \mathrm{~F}\) (a megabecquerel is a unit of radioactivity). The patient then undergoes a PET scan (positron emission tomography) to detect areas of metabolic activity indicative of cancer. After \(174 \mathrm{~min}\), one-third of the original dose remains in the body. a. Write a function of the form \(Q(t)=Q_{0} e^{-k t}\) to model the radioactivity level \(Q(t)\) of fluorine- 18 at a time \(t\) minutes after the initial dose. b. What is the half-life of \({ }^{18} \mathrm{~F}\) ?

Solve the equation. Write the solution set with the exact solutions. Also give approximate solutions to 4 decimal places if necessary. \(\log (p+17)=4.1\)

Find the difference quotient \(\frac{f(x+h)-f(x)}{h} .\) Write the answers in factored form. $$f(x)=e^{x}$$
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