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Chapter 2: Problem 57

The function \(f\) is not one-to-one on the given domain but is one-to-one on the restricted domain (the second interval). Find the inverse of the one-to- one function and give its domain. Sketch the graph of \(f\) on the restricted domain and the graph of \(f^{-1}\) on the same coordinate axes. $$ f(x)=4 x^{2}+2,(-\infty, \infty) ;[0, \infty) $$

Short Answer

Expert verified
The inverse is \(f^{-1}(x) = \sqrt{\frac{x-2}{4}}\) with domain \([2, \infty)\).

Step by step solution

01

Identify the Function and Domain

The function given is \( f(x) = 4x^2 + 2 \) and it is not one-to-one over the domain \((- Alibaba, \infty)\). However, it is one-to-one on the restricted domain \([0, \infty)\).
02

Understand One-to-One Restriction

A function is one-to-one if every y-value is matched with exactly one x-value. By restricting the domain to \([0, \infty)\), we avoid the issue of \(x^2\) being symmetrical, thus allowing inversion.
03

Set up the Equation for Inversion

To find the inverse, set \(y = f(x)\). Thus, \(y = 4x^2 + 2\). Solve for \(x\) in terms of \(y\) to find \(f^{-1}(y)\).
04

Solve the Equation for \(x\)

Subtract 2 from both sides: \(y - 2 = 4x^2\). Divide by 4: \(x^2 = \frac{y - 2}{4}\). Take the square root: \(x = \sqrt{\frac{y - 2}{4}}\), but since \(x\) must be non-negative over the domain \([0, \infty)\), the inverse is \(f^{-1}(y) = \sqrt{\frac{y - 2}{4}}\).
05

Determine the Inverse's Domain

The domain of \(f^{-1}\) corresponds to the range of \(f\) over the restricted domain, which is \(f(x) = 4x^2 + 2\). For \(x \geq 0\), \(f(x)\) ranges from 2 to \(\infty\). Thus, the domain of \(f^{-1}\) is \([2, \infty)\).
06

Graph \(f(x)\) on Restricted Domain

Graph \(f(x) = 4x^2 + 2\) starting from \(x = 0\). The curve is a parabola opening upwards, starting at point (0,2) and extending towards positive \(y\) as \(x\) increases.
07

Graph \(f^{-1}(x)\) on the Same Axes

Graph \(f^{-1}(x) = \sqrt{\frac{x-2}{4}}\) starting from \(x=2\). This graph will be the reflection of \(f(x)\) across the line \(y=x\), starting at point (2,0) and moving upward to the right.

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Key Concepts

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

One-to-one Function
A function is called "one-to-one" when each output value corresponds to exactly one input value. This means that no two different inputs will ever produce the same output. It鈥檚 an essential characteristic when finding an inverse function, because if a function is not one-to-one, its inverse would not be well-defined. When dealing with a quadratic function like \( f(x) = 4x^2 + 2 \), it is generally not one-to-one over its entire domain because squaring creates symmetry, meaning the function will repeat values. Recognizing this, we often restrict the domain to ensure the function is one-to-one, thus enabling us to find a valid inverse.
Restricted Domain
When a function isn't one-to-one over its entire domain, we can restrict it to a subset where it behaves as a one-to-one function. For the function \( f(x) = 4x^2 + 2 \), we restrict the domain to \([0, \infty)\). This restriction is chosen because for \( x \geq 0 \), \( 4x^2 \) will increase as \( x \) increases, ensuring each \( y \) value comes from a unique \( x \) value. The restricted domain eliminates the problem of symmetry found in the quadratic function, helping us avoid confusion caused by repeated values and making it a valid candidate for finding an inverse.
Graphing Functions
Understanding the graphical representation of functions helps in grasping their behavior visually.
  • To graph \( f(x) = 4x^2 + 2 \) over the restricted domain \([0, \infty)\), start plotting from the point \((0,2)\), where the parabola begins at the vertex and opens upwards. The graph will continue extending to the right as \( x \) increases.
  • For the inverse function \( f^{-1}(x) = \sqrt{\frac{x - 2}{4}} \): Begin plotting from \((2,0)\), reflecting the original function across the line \( y = x \). This helps in visualizing how the inverse flips the roles of \( x \) and \( y \), leading to the inverse curve rising towards the right.
This dual plotting helps you see how each function and its inverse relate to each other geometrically.
Square Root Function
The inverse of many functions involving squares becomes a type of square root function, as seen here with \( f^{-1}(x) = \sqrt{\frac{x - 2}{4}} \). Square root functions have distinct properties:
  • The square root function is defined only for non-negative inputs, reflecting the range limit of the original function defined over \( [0, \infty) \).
  • This particular form, \( \sqrt{\frac{x - 2}{4}} \), arises from reversing the squaring process. By isolating \( x \), subtracting constant terms, and dividing, we arrive at the point where taking the square root gives us the inverse.
  • Square root functions typically have a gentle rise, as they start slowly but increase more rapidly past a certain point, mirroring the growth from the point \( (2,0) \) as \( x \) grows larger.
Recognizing these patterns is vital for understanding both the algebraic and graphical behaviors of inverse operations.

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Most popular questions from this chapter

The given function \(f\) is one-toone. Without finding \(f^{-1}\) find its domain and range. $$ f(x)=4+\sqrt{x} $$

Proceed as in Example 3 and verify that the inverse of the one-to-one function \(f\) is the function \(g\) by showing \(f(g(x))=x\) and \(g(f(x))=x\). $$ f(x)=\frac{x-3}{x+1} ; g(x)=\frac{x+3}{1-x} $$

Consider the semicircle whose equation is \(f(x)=\sqrt{1-x^{2}} .\) Discuss: How can the derivative \(f(x)\) be found using only the geometric fact that the radius of a circle is perpendicular to the tangent line at a point \((x, y)\) on the circle?

The function \(f\) is not one-to-one on the given domain but is one-to-one on the restricted domain (the second interval). Find the inverse of the one-to- one function and give its domain. Sketch the graph of \(f\) on the restricted domain and the graph of \(f^{-1}\) on the same coordinate axes. $$ f(x)=\sqrt{1-x^{2}},[-1,1] ;[0,1] $$

In Problems 35 and 36 , sketch the graph the given piece wise-defined function \(f\) for \(c=-2,-1,1,2\). $$ f(x)=\left\\{\begin{array}{ll} x^{2}+c, & x<0 \\ -x^{2}+c & x \geq 0 \end{array}\right. $$
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