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Provide additional review on the inverse function concept. The exercises do not explicilly involve logarithmic functions (but may be helpful to you because we've defined logarithmic functions in terms of inverse functions). Let \(f(x)=x^{3}+5 x+1 .\) Given that \(f^{-1}\) exists and 6 is in the domain of \(f^{-1}\), evaluate \(f\left[f^{-1}(6)\right]\)

Short Answer

Expert verified
The result is 6 because \( f(f^{-1}(x)) = x \) for any \( x \) in the domain of \( f^{-1} \).

Step by step solution

01

Understand the Function and Its Inverse

We are given the function \( f(x) = x^3 + 5x + 1 \) and need to evaluate \( f[f^{-1}(6)] \). The expression \( f^{-1}(6) \) represents the input \( x \) for which \( f(x) = 6 \). The property of inverse functions is that \( f(f^{-1}(x)) = x \) for any \( x \) in the domain of \( f^{-1} \).
02

Evaluate f[f^{-1}(6)] Using Inverse Function Property

Since \( f^{-1} \) is the inverse of \( f \), then for any value \( y \) in the domain of the inverse, \( f(f^{-1}(y)) = y \). Thus, \( f[f^{-1}(6)] = 6 \) because \( 6 \) is within the domain of \( f^{-1} \).

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Key Concepts

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

Function Composition
Function composition is a fundamental concept in mathematics. It involves combining two functions to form a new function. When we talk about function composition, we think about applying one function to the result of another. It's like stacking tasks or instructions in a sequence.
Notationally, if we have two functions, say \( f(x) \) and \( g(x) \), composing \( f \) with \( g \) is written as \( f(g(x)) \). This means that you first apply \( g \) to \( x \) and then apply \( f \) to the result of \( g(x) \).
  • **Order Matters:** The order in which you compose functions is significant; \( f(g(x)) \) is generally not the same as \( g(f(x)) \).
  • **Identity Role:** The identity function acts as a neutral element, meaning if you compose a function with the identity function, it remains unchanged.
In the context of inverse functions, composition plays a special role: composing a function with its inverse gives you the original input. That is, \( f(f^{-1}(x)) = x \) and \( f^{-1}(f(x)) = x \) whenever the compositions are defined.
Function Domain
The domain of a function is the complete set of possible values of the independent variable, usually denoted as \( x \). In simpler terms, it tells us all the possible inputs a function can handle without breaking any rules.
Understanding the domain is crucial when dealing with functions and especially important when considering inverse functions.
  • For a function \( f(x) \), we often write its domain as \( D(f) \).
  • The domain of an inverse function, \( f^{-1} \), is the range (set of possible outputs) of the original function \( f(x) \).
  • Inverse functions reverse the roles of input and output, so the domain and range of the original function swap for its inverse.
When solving or simplifying expressions involving inverse functions, knowledge of domains ensures those expressions are both valid and meaningful. In our example, since \( 6 \) is within the domain of \( f^{-1} \), we can confidently apply the inverse and compute \( f[f^{-1}(6)] = 6 \).
Cubic Function
A cubic function is a polynomial of degree three, which means its highest power of \( x \) is three. The general form of a cubic function is \( f(x) = ax^3 + bx^2 + cx + d \), where \( a, b, c, \) and \( d \) are constants and \( a eq 0 \).
The function \( f(x) = x^3 + 5x + 1 \) is a specific instance of a cubic function.
  • **Non-linear Behavior:** Cubic functions are non-linear, having an S-shaped curve when graphed, with possible inflection points where the curve changes from concave to convex or vice versa.
  • **Roots & Turning Points:** A cubic function can have up to three real roots and typically has one or two turning points.
  • **Behavior at Infinity:** As \( x \) approaches positive or negative infinity, the function itself will behave similarly, increasing or decreasing without bounds.
Cubic functions are crucial in understanding polynomial behaviors because they introduce complexity beyond linear and quadratic functions, offering rich behavior for study.

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Most popular questions from this chapter

(a) Use a graphing utility to estimate the root(s) of the equation to the nearest one-tenth (as in Example 6). (b) Solve the given equation algebraically by first rewriting it in logarithmic form. Give two forms for each answer: an exact expression and a calculator approximation rounded to three decimal places. Check to see that each result is consistent with the graphical estimate obtained in part (a). $$e^{2 t+3}=10$$

The intensity of the sounds that the human ear can detect varies over a very wide range of values. For instance, a whisper from 1 meter away has an intensity of approximately \(10^{-10}\) watts per square meter \(\left(\mathrm{W} / \mathrm{m}^{2}\right)\), whereas, from a distance of 50 meters, the intensity of a launch of the Space Shuttle is approximately \(10^{8} \mathrm{W} / \mathrm{m}^{2} .\) For a sound with intensity \(I\), the sound level \(\beta\) is defined by $$ \beta=10 \log _{10}\left(I / I_{0}\right) $$ where the constant \(I_{0}\) is the sound intensity of a barely audible sound at the threshold of hearing. The units for the sound level \(\beta\) are decibels, abbreviated dB. (a) Solve the equation \(\beta=10 \log _{10}\left(1 / I_{0}\right)\) for \(I\) by first dividing by 10 and then converting to exponential form. (b) The sound level for a power lawnmower is \(\beta=100 \mathrm{db}\). and that for a cat purring is \(\beta=10\) db. Use your result in part (a) to determine how many times more intense is the power mower sound than the cat's purring.

(a) Graph the two functions \(f(x)=(\ln x) /(\ln 3)\) and \(g(x)=\ln x-\ln 3 .\) (Use a viewing rectangle in which \(x\) extends from 0 to 10 and \(y\) extends from \(-5 \text { to } 5 .)\) Why aren't the two graphs identical? That is, doesn't one of the basic log identities say that \((\ln a) /(\ln b)=\ln a-\ln b ?\) (b) Your picture in part (a) indicates that $$\frac{\ln x}{\ln 3}>\ln x-\ln 3 \quad(0< x \leq 10)$$ Find a viewing rectangle in which \((\ln x) /(\ln 3) \leq \ln x-\ln 3.\) (c) Use the picture that you obtain in part (b) to estimate the value of \(x\) for which \((\ln x) /(\ln 3)=\ln x-\ln 3.\) (d) Solve the equation \((\ln x) /(\ln 3)=\ln x-\ln 3\) algebraically and use the result to check your estimate in part (c).

Solve the inequalities. Where appropriate, give an exact answer as well as a decimal approximation. $$3 \log _{10}(4 x+3)<1$$

Decide which of the following properties apply to each function. (More than one property may apply to a function.)A. The function is increasing for \(-\infty

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