/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 36 Find the inverse function of \(f... [FREE SOLUTION] | 91影视

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Chapter 4: Problem 36

Find the inverse function of \(f.\) \(f(x)=-x^{3}+2\)

Short Answer

Expert verified
The inverse function is \( f^{-1}(x) = \sqrt[3]{2 - x} \).

Step by step solution

01

Set the function equal to y

First, we represent the function as an equation in terms of y. Replace \( f(x) \) with \( y \): \[ y = -x^3 + 2 \]
02

Solve for x in terms of y

To find the inverse function, we need to express \( x \) in terms of \( y \). Start by isolating the term with \( x \): \[ y - 2 = -x^3 \]. Dividing both sides by -1 gives \[ x^3 = 2 - y \]. Finally, solve for \( x \) by taking the cube root of both sides: \[ x = \sqrt[3]{2 - y} \].
03

Swap x and y

To express \( x \) in terms of \( y \), we now swap the roles of \( x \) and \( y \). This gives us the inverse function: \[ y = \sqrt[3]{2 - x} \]. So, the inverse function can be written as \( f^{-1}(x) = \sqrt[3]{2 - x} \).
04

Write the Inverse Function

Finally, write the expression for the inverse function we found in the previous step: \[ f^{-1}(x) = \sqrt[3]{2 - x} \] This is the inverse of the original function.

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Key Concepts

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

Cubic Functions
Cubic functions are a type of polynomial function where the highest degree of the variable is three. These functions have the general form \[ f(x) = ax^3 + bx^2 + cx + d \]where \(a\), \(b\), \(c\), and \(d\) are constants, and \(a eq 0\). The specific function we are dealing with, \( f(x) = -x^3 + 2 \), is a simplified cubic function where the coefficients \(b\) and \(c\) are zero.
  • Cubic functions can have up to three real roots, or solutions, due to their high degree.
  • Their graphs possess a characteristic 'S' shape and can cross the x-axis up to three times.
  • These functions often model situations in physics and engineering due to their complex yet smooth transitions.
Understanding cubic functions is crucial for analyzing inverse functions involving cubics, as they require algebraic manipulation of a cubic term.
Function Notation
Function notation is a compact and elegant way to depict relationships between variables. Instead of expressing functions in a lengthy sentence, we use function notation to simplify this process.
  • The notation \(f(x)\) represents a function named \(f\) where \(x\) is the input variable.
  • The output, or the result of substituting \(x\) in the function, is usually dependent on a formula or rule defined for \(f\).
  • In our case, \(f(x) = -x^3 + 2\) shows that to find the function's output, you substitute \(x\) into \(-x^3 + 2\).
Function notation not only clarifies the function's output for given inputs but also lays the foundation for expressing inverse functions. When working with inverses, where \(f(x)\) becomes \(f^{-1}(x)\), it is important to keep track of which variable represents the input and which represents the output.
Solving Equations
Solving equations is a fundamental skill in algebra that involves finding the value of variables that satisfy a given equation. For our exercise, solving equations involved manipulating the equation to find the inverse function.
  • First, replace \(f(x)\) with \(y\), creating the equation \(y = -x^3 + 2\).
  • The goal is to solve for \(x\) in terms of \(y\). This means rewriting the equation so that \(x\) is isolated on one side.
  • We do this by isolating terms, changing signs, and taking cube roots: \(x^3 = 2 - y\), which simplifies to \(x = \sqrt[3]{2 - y}\).
The process of solving such equations often requires a good understanding of algebraic manipulation and inverse operations, like finding a cube root to counter a cube.
Algebraic Manipulation
Algebraic manipulation refers to the process of rearranging and simplifying equations to solve for a variable or to express the relationship between variables differently. This skill becomes particularly useful when finding inverse functions.
  • In our example, we needed to manipulate \(y = -x^3 + 2\) to solve for \(x\). This involved moving terms across an equal sign and changing operations, such as moving from addition to subtraction.
  • Taking the cube root is a form of algebraic manipulation used to "undo" the cube in \(x^3\).
  • Mastering algebraic manipulation allows you to swap variables, as we did by exchanging \(x\) and \(y\), resulting in a new inverse function representation.
Through practice, algebraic manipulation can transform how you solve complex equations, allowing for clarity and precision in mathematics, particularly with inverse functions.

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Most popular questions from this chapter

Cooling A jar of boiling water at \(212^{\circ} \mathrm{F}\) is set on a table in a room with a temperature of \(72^{\circ} \mathrm{F}\). If \(T(t)\) represents the temperature of the water after \(t\) hours, graph \(T(t)\) and determine which function best models the situation. (1) \(T(t)=212-50 t\) (2) \(T(t)=140 e^{-t}+72\) (3) \(T(t)=212 e^{-t}\) (4) \(T(t)=72+10 \ln (140 t+1)\)

Exer. \(87-88:\) Graph \(f\) and \(g\) on the same coordinate plane, and estimate the solution of the inequality \(f(x) \geq g(x)\) $$f(x)=x \ln |x|: \quad g(x)=0.15 e^{x}$$

Computer chips For manufacturers of computer chips, it is important to consider the fraction \(F\) of chips that will fail after \(t\) years of service. This fraction can sometimes be approximated by the formula \(F=1-e^{-c t},\) where \(c\) is a positive constant. (a) How does the value of \(c\) affect the reliability of a chip? (b) If \(c=0.125,\) after how many years will \(35 \%\) of the chips have failed?

Graph \(f\) on the interval [0.2, 16]. (a) Estimate the intervals where \(f\) is increasing or is decreasing. (b) Estimate the maximum and minimum values of \(f\) on \([0.2,16]\). $$f(x)=2 \log 2 x-1.5 x+0.1 x^{2}$$

Vertical wind shear Refer to Exercises \(67-68\) in Section \(2.3 .\) If \(v_{0}\) is the wind speed at height \(h_{0}\) and if \(v_{1}\) is the wind speed at height \(h_{1},\) then the vertical wind shear can be described by the equation $$ \frac{v_{0}}{v_{1}}=\left(\frac{h_{0}}{h_{1}}\right)^{P} $$ where \(P\) is a constant. During a one-year period in Montreal, the maximum vertical wind shear occurred when the winds at the 200 -foot level were \(25 \mathrm{mi} / \mathrm{hr}\) while the winds at the 35 -foot level were \(6 \mathrm{mi} / \mathrm{hr}\). Find \(P\) for these conditions.
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