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Chapter 3: Problem 95

A function \(f\) and its domain are given. Graph \(x=f^{-1}(y)\) in the \(y x\) -plane. (Do not attempt to find a formula for \(f^{-1}\). Recall that a procedure for graphing an inverse function has been described in Section 1.5.) Find the equation of the tangent line to the graph of \(x=f^{-1}(y)\) at \(\left(2, f^{-1}(2)\right)\). Add the graph of the tangent line to your plot. $$ f(x)=\sin ^{2}(x)-\cos (x)+2.6,0 \leq x \leq 2 $$

Short Answer

Expert verified
Reflect the graph of \( f \) to plot \( x = f^{-1}(y) \), evaluate \( f(2) \), determine \( f'(2) \), and use these to sketch the tangent line.

Step by step solution

01

Understand the Problem Statement

We need to graph the inverse of the function \( f(x) = \sin^2(x) - \cos(x) + 2.6 \) for the domain \( 0 \leq x \leq 2 \). Additionally, we must find and graph the tangent line to this inverse function's graph at the point \( (2, f^{-1}(2)) \).
02

Review Inverse Function Properties

The graph of an inverse function \( f^{-1} \) is obtained by reflecting the graph of \( f \) across the line \( y = x \). Therefore, we plot \( x = f^{-1}(y) \) in the \( yx \)-plane by reflecting the graph of \( y = f(x) \) across this line.
03

Graph the Function \( y = f(x) \)

Plot \( y = \sin^2(x) - \cos(x) + 2.6 \) over \( 0 \leq x \leq 2 \) to visualize how the function behaves, as it will help to manually sketch the inverse.
04

Reflect the Graph to Obtain \( x = f^{-1}(y) \)

Reflect the points of \( y = f(x) \) across the line \( y = x \) to get the graph \( x = f^{-1}(y) \). If a point \( (a,b) \) is on \( f \), then \( (b,a) \) will be on \( f^{-1} \).
05

Determine \( f(2) \)

Evaluate \( f(2) = \sin^2(2) - \cos(2) + 2.6 \) to find the y-coordinate for which we need the tangent line. Calculate this numerically.
06

Differentiate \( f(x) \)

Find the derivative \( f'(x) \) to help us determine the slope of the tangent to \( f^{-1} \). Given \( f(x) = \sin^2(x) - \cos(x) + 2.6 \), apply the chain rule and differentiate: \( f'(x) = 2\sin(x)\cos(x) + \sin(x) = \sin(2x) + \sin(x) \).
07

Calculate the Slope of the Tangent Line

The slope of the tangent line to \( x = f^{-1}(y) \) at a point where \( f(x) = y \) is \( \frac{1}{f'(x)} \). Using \( x = 2 \), calculate \( f'(2) \), and find the reciprocal for the slope at that point.
08

Write the Equation of the Tangent Line

For the point \( (f(2), 2) \) in \( yx \)-plane, the tangent line equation is: \( y - 2 = \frac{1}{f'(2)} (x - f(2)) \). Substitute the values to get the equation.

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Key Concepts

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

Function Graphing
When graphing a function, the goal is to visually represent the relationship between inputs and outputs. In this case, we start by graphing the original function \( f(x) = \sin^2(x) - \cos(x) + 2.6 \) over the specified domain, \( 0 \leq x \leq 2 \). This function combines trigonometric components, making its graph potentially more complex. To better understand it:
  • Plot a series of points within the domain to see how \( y = f(x) \) behaves.
  • Notice that the function involves both sine and cosine, which are periodic.
  • Adding a constant, 2.6, shifts the entire function upwards.
By plotting these points and connecting them smoothly, the graph gives us a visual representation of how \( f(x) \) changes with \( x \).
The next task is constructing the inverse graph, \( x = f^{-1}(y) \). Consider that finding the exact formula for the inverse can be complex, so graphing is often done by reflection.
Tangent Line
A tangent line is a straight line that touches a curve at one single point and has the same slope as the curve at that point. This line represents the curve's immediate direction and rate of change at that point.
To find the tangent line for \( x = f^{-1}(y) \) at the point \((2, f^{-1}(2))\), follow these steps:
  • First, identify the corresponding point on the graph \((f(2), 2)\) in the \( yx \)-plane. Solve \( f(2) = \sin^2(2) - \cos(2) + 2.6 \) for a specific value, providing the y-coordinate of the inverse location.
  • Use this value to establish the exact point where the tangent touches the graph in the inverse function setting.
  • The equation of the tangent will follow the general formula: \( y - b = m \cdot (x - a) \), where \((a, b)\) is the point of tangency and \( m \) is the slope.
Establish the tangent equation at this intersection to graph it alongside the inverse function, giving a complete view of changes around that point.
Differentiation
Differentiation is a process in calculus used to determine the rate at which a function is changing at any point on its graph. This rate is known as the derivative. For our function, \( f(x) = \sin^2(x) - \cos(x) + 2.6 \), the derivative \( f'(x) \) provides vital insight into how steep the function is at different points.
Applying differentiation:
  • Use the chain rule to differentiate \( \sin^2(x) \), resulting in \( 2\sin(x)\cos(x) = \sin(2x) \).
  • Differentiate \( -\cos(x) \) using the basic derivative, giving \( \sin(x) \).
  • Combine these results: \( f'(x) = \sin(2x) + \sin(x) \).
The derivative is crucial for determining the slope of the tangent line to the inverse function. Since the slope of a tangent in \( x = f^{-1}(y) \) is the reciprocal of \( f'(x) \), calculate \( f'(2) \) and use its reciprocal for the exact slope value of the tangent line at that specific point.
Domain and Range
Understanding a function's domain and range is essential to correctly characterizing where it applies and how its values behave. For the function \( f(x) = \sin^2(x) - \cos(x) + 2.6 \), the domain is clearly delineated as \( 0 \leq x \leq 2 \).
Breaking down these concepts:
  • The domain specifies all possible input values (x-values) the function accepts.
  • The range refers to all possible outputs (y-values) the function can produce.
  • By examining \( f \) over its domain, we determine its range, affecting what the inverse graph can cover.
For \( x = f^{-1}(y) \), roles of domain and range invert compared to \( f \). The former range now serves as the domain for the inverse, and vice versa. By confirming the range of \( f \) within its domain, one gains clearer insight into what \( f^{-1} \) can yield, aiding in accurately plotting its graph.

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Most popular questions from this chapter

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