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Chapter 1: Problem 20

Suppose a function \(f(x)\) has an inverse function, \(f^{-1}(x)\). a) Determine \(f^{-1}(5)\) if \(f(17)=5\). b) Determine \(f(-2)\) if \(f^{-1}(\sqrt{3})=-2\). c) Determine the value of \(a\) if \(f^{-1}(a)=1\) and \(f(x)=2 x^{2}+5 x+3, x \geq-1.25\).

Short Answer

Expert verified
\(f^{-1}(5) = 17\), \(f(-2) = \sqrt{3}\), and \(a = 10\).

Step by step solution

01

Understanding the Inverse Function Property

The inverse function, denoted as \(f^{-1}(x)\), reverses the effect of the original function \(f(x)\). This means that if \(y = f(x)\), then \(x = f^{-1}(y)\).
02

Determine \(f^{-1}(5)\) given \(f(17) = 5\)

Since \(f(17) = 5\), by the property of inverse functions, \(f^{-1}(5) = 17\).
03

Determine \(f(-2)\) given \(f^{-1}(\sqrt{3}) = -2\)

Given \(f^{-1}(\sqrt{3}) = -2\), by the property of inverse functions, this means \(f(-2) = \sqrt{3}\).
04

Determine value of \(a\) such that \(f^{-1}(a) = 1\)

Given \(f(x) = 2x^2 + 5x + 3\), we need to find \(a\) such that \(f^{-1}(a) = 1\). This implies \(f(1) = a\). Calculate \(f(1)\): \[ f(1) = 2(1)^2 + 5(1) + 3 = 2 + 5 + 3 = 10. \] Therefore, \(a = 10\).

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Key Concepts

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

Inverse Function Property
When dealing with inverse functions, the core idea is that they reverse the operation of the original function. If the original function is denoted as \(f(x)\), then the inverse function is represented as \(f^{-1}(x)\). The key relationship here is that if \(y = f(x)\), then \(x = f^{-1}(y)\). This property helps us solve many problems where we need to find values of the inverse function.

For example, if we know that \(f(17) = 5\), by applying the inverse function property, we can see that \(f^{-1}(5) = 17\). This relationship always holds true and is fundamental to understanding and working with inverse functions effectively.
Determining Inverse Function Values
Once you grasp the inverse function property, you can determine specific values of the inverse function. Let's look at some practical examples.

Given the following problems:

  • Find \(f^{-1}(5)\) if \(f(17) = 5\).
  • Find \(f(-2)\) if \(f^{-1}(\text{鈭殅3) = -2\).
By the inverse function property:

  • Since \(f(17) = 5\), it means \(f^{-1}(5) = 17\).
  • Since \(f^{-1}(\text{鈭殅3) = -2\), it implies \(f(-2) = \text{鈭殅3\).
As long as you remember the relationship, it's straightforward to determine the needed values.
Precalculus Problems
In precalculus, solving for inverses often involves dealing with more complex functions. Let's tackle an example using the function \(f(x) = 2x^2 + 5x + 3\) with \( x \geq -1.25\).

We need to find the value of \(a\) such that \(f^{-1}(a) = 1\). According to the inverse function property, this means \(f(1) = a\).

Calculate \(f(1)\): \[f(1) = 2(1)^2 + 5(1) + 3 = 2 + 5 + 3 = 10.\] Therefore, \(a = 10\).

This demonstrates how we can use the given function to find specific values and reinforces the importance of understanding the relationship between a function and its inverse.

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Most popular questions from this chapter

An object falling in a vacuum is affected only by the gravitational force. An equation that can model a free-falling object on Earth is \(d=-4.9 t^{2},\) where \(d\) is the distance travelled, in metres, and \(t\) is the time, in seconds. An object free falling on the moon can be modelled by the equation \(d=-1.6 t^{2}\) a) Sketch the graph of each function. b) Compare each function equation to the base function \(d=t^{2}\)

The speed of a vehicle the moment the brakes are applied can be determined by its skid marks. The length, \(D\), in feet, of the skid mark is related to the speed, \(S\) in miles per hour, of the vehicle before braking by the function \(D=\frac{1}{30 f n} S^{2},\) where \(f\) is the drag factor of the road surface and \(n\) is the braking efficiency as a decimal. Suppose the braking efficiency is \(100 \%\) or 1 a) Sketch the graph of the length of the skid mark as a function of speed for a drag factor of \(1,\) or \(D=\frac{1}{30} S^{2}\) b) The drag factor for asphalt is \(0.9,\) for gravel is \(0.8,\) for snow is \(0.55,\) and for ice is 0.25. Compare the graphs of the functions for these drag factors to the graph in part a).

Given the function \(y=f(x),\) write the equation of the form \(y-k=a f(b(x-h))\) that would result from each combination of transformations. a) a vertical stretch about the \(x\) -axis by a factor of \(3,\) a reflection in the \(x\) -axis, a horizontal translation of 4 units to the left, and a vertical translation of 5 units down b) a horizontal stretch about the \(y\) -axis by a factor of \(\frac{1}{3},\) a vertical stretch about the \(x\) -axis by a factor of \(\frac{3}{4},\) a reflection in both the \(x\) -axis and the \(y\) -axis, and a translation of 6 units to the right and 2 units up

Consider the function \(f(x)=(x+4)(x-3)\) Without graphing, determine the zeros of the function after each transformation. a) \(y=4 f(x)\) b) \(y=f(-x)\) c) \(y=f\left(\frac{1}{2} x\right)\) d) \(y=f(2 x)\)

For each function, state the values of \(h\) and \(k,\) the parameters that represent the horizontal and vertical translations applied to \(y=f(x)\). a) \(y-5=f(x)\) b) \(y=f(x)-4\) \(y=f(x+1)\) d) \(y+3=f(x-7)\) e) \(y=f(x+2)+4\)
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