Problem 62
Question
You will explore some functions and their inverses together with their derivatives and linear approximating functions at specified points. Perform the following steps using your CAS: a. Plot the function \(y=f(x)\) together with its derivative over the given interval. Explain why you know that \(f\) is one-to-one over the interval. b. Solve the equation \(y=f(x)\) for \(x\) as a function of \(y,\) and name the resulting inverse function \(g\) c. Find the equation for the tangent line to \(f\) at the specified point \(\left(x_{0}, f\left(x_{0}\right)\right)\) d. Find the equation for the tangent line to \(g\) at the point \(\left(f\left(x_{0}\right), x_{0}\right)\) located symmetrically across the \(45^{\circ}\) line \(y=x\) (which is the graph of the identity function). Use Theorem 1 to find the slope of this tangent line. e. Plot the functions \(f\) and \(g,\) the identity, the two tangent lines, and the line segment joining the points \(\left(x_{0}, f\left(x_{0}\right)\right)\) and \(\left(f\left(x_{0}\right), x_{0}\right) .\) Discuss the symmetries you see across the main diagonal. $$y=\frac{3 x+2}{2 x-11}, \quad-2 \leq x \leq 2, \quad x_{0}=1 / 2$$
Step-by-Step Solution
Verified Answer
The function is one-to-one as its derivative is non-zero over the interval. The inverse is \(g(y) = \frac{11y + 2}{2y - 3}\). Tangent lines illustrate symmetry across \(y = x\).
1Step 1: Plot the Function and Derivative
Plot the function \(y=\frac{3x+2}{2x-11}\) over the interval \(-2 \leq x \leq 2\) along with its derivative. The derivative is calculated as \(f'(x) = \frac{-37}{(2x - 11)^2}\). Since \(f'(x)\) is non-zero and does not change sign over the interval, it confirms that \(f\) is one-to-one there.
2Step 2: Solve for the Inverse Function
To find the inverse, solve \(y = \frac{3x+2}{2x-11}\) for \(x\). Rearranging gives \(x = \frac{11y + 2}{2y - 3}\), so the inverse function is \(g(y) = \frac{11y + 2}{2y - 3}\).
3Step 3: Find the Tangent Line to f at the Specified Point
The tangent line to \(f\) at \(x_0 = \frac{1}{2}\) is found using the point \(\left(\frac{1}{2}, f\left(\frac{1}{2}\right)\right) = \left(\frac{1}{2}, -2.5\right)\). Calculate the slope as \(f'(\frac{1}{2}) = \frac{-37}{\left(2\left(\frac{1}{2}\right) - 11\right)^2} = \frac{-37}{100}\). Hence, the equation is \(y + 2.5 = \frac{-37}{100}\left(x - \frac{1}{2}\right)\).
4Step 4: Find the Tangent Line to g at the Symmetric Point
The tangent line to \(g\) at \(\left(-2.5, \frac{1}{2}\right)\) is calculated by finding the slope using Theorem 1: If \(f'\left(x_0\right)\) is \(-37/100\), then \(g'\left(y_0\right) = -1/f'(x_0) = 100/37\). The equation thus is \(x = \frac{100}{37}(y + 2.5) + \frac{1}{2}\).
5Step 5: Plot and Discuss the Symmetries
Plot \(f\), \(g\), the identity line \(y=x\), the tangent lines from Steps 3 and 4, and the line segment joining \((\frac{1}{2}, -2.5)\) and \((-2.5, \frac{1}{2})\). The plots demonstrate symmetry about the line \(y = x\), showing how \(f\) and its inverse \(g\) and their tangents reflect across this diagonal.
Key Concepts
DerivativeOne-to-one FunctionTangent LineMathematical Symmetries
Derivative
The derivative of a function provides crucial insights into its behavior. When we differentiate a function, we obtain another function that describes how the original function changes at any given point. For the function in our exercise, \[ f(x) = \frac{3x+2}{2x-11} \], we find its derivative as \[ f'(x) = \frac{-37}{(2x - 11)^2} \]. The negative sign indicates that the function is decreasing over the interval where it is defined.
Why is the derivative important here? Because its behavior affects whether the function is one-to-one. Since \( f'(x) \) is non-zero and does not change sign in the interval, \[-2 \leq x \leq 2\], it confirms that \( f \) is consistently decreasing, making it a one-to-one function. A strict monotonic (always increasing or decreasing) function over an interval ensures that every y-value corresponds to one and only one x-value, making finding an inverse possible.
Why is the derivative important here? Because its behavior affects whether the function is one-to-one. Since \( f'(x) \) is non-zero and does not change sign in the interval, \[-2 \leq x \leq 2\], it confirms that \( f \) is consistently decreasing, making it a one-to-one function. A strict monotonic (always increasing or decreasing) function over an interval ensures that every y-value corresponds to one and only one x-value, making finding an inverse possible.
One-to-one Function
A one-to-one function is an essential concept in calculus and algebra, particularly when dealing with inverse functions. A function is one-to-one if every element of the range is paired with a unique element from the domain.
In simpler terms, if you have two inputs, \( x_1 \) and \( x_2 \), resulting in the same output (i.e., \( f(x_1) = f(x_2) \)), then it must mean that \( x_1 = x_2 \). This property allows a function to have an inverse because each output corresponds to a single input.
In our context, since the derivative \( f'(x) \) does not change sign within \[-2 \leq x \leq 2\], \( f \) is a strictly monotone decreasing function. This guarantees the function is one-to-one over its defined interval, which simply means the inverse can be calculated smoothly.
In simpler terms, if you have two inputs, \( x_1 \) and \( x_2 \), resulting in the same output (i.e., \( f(x_1) = f(x_2) \)), then it must mean that \( x_1 = x_2 \). This property allows a function to have an inverse because each output corresponds to a single input.
In our context, since the derivative \( f'(x) \) does not change sign within \[-2 \leq x \leq 2\], \( f \) is a strictly monotone decreasing function. This guarantees the function is one-to-one over its defined interval, which simply means the inverse can be calculated smoothly.
Tangent Line
The tangent line to a curve at a given point is the straight line that 'just touches' the curve at that point. It gives the best linear approximation of the function near that point. To find the equation of the tangent line, one needs to know:
Similarly, for the inverse function \( g \), the tangent line at the symmetric point is found using Theorem 1, which relates the derivatives of inverse functions and provides the equation for its tangent as \[ x = \frac{100}{37}(y + 2.5) + \frac{1}{2} \]. This illustrates how closely related the functions and their inverses are with respect to their slopes.
- The slope of the function at the point, given by the derivative.
- The coordinates of the point on the function.
Similarly, for the inverse function \( g \), the tangent line at the symmetric point is found using Theorem 1, which relates the derivatives of inverse functions and provides the equation for its tangent as \[ x = \frac{100}{37}(y + 2.5) + \frac{1}{2} \]. This illustrates how closely related the functions and their inverses are with respect to their slopes.
Mathematical Symmetries
Mathematical symmetries are beautiful properties that make functions and their visuals fascinating. Symmetry primarily means that one part of a shape or object is a mirror image of another. When observing functions and their inverses, understanding symmetry is crucial.
In our exercise, the symmetry can be seen in the reflection of the function \( f \) and its inverse \( g \) across the identity line \( y = x \). When you draw both functions, \( f(x) \) and \( g(y) \), along with this line, there's a beautiful mirroring effect about the line \( y = x \).
The tangent lines and their intersection points showcase this symmetry too. The line segment connecting the coordinates \((\frac{1}{2}, -2.5)\) and \(( -2.5, \frac{1}{2} )\) lies perfectly across the main diagonal, emphasizing how inverses function in reflecting across these axes. Observing these symmetries not only helps in understanding the mathematics but also reinforces the significance of the \( y = x \) line as a critical feature of inverse relationships.
In our exercise, the symmetry can be seen in the reflection of the function \( f \) and its inverse \( g \) across the identity line \( y = x \). When you draw both functions, \( f(x) \) and \( g(y) \), along with this line, there's a beautiful mirroring effect about the line \( y = x \).
The tangent lines and their intersection points showcase this symmetry too. The line segment connecting the coordinates \((\frac{1}{2}, -2.5)\) and \(( -2.5, \frac{1}{2} )\) lies perfectly across the main diagonal, emphasizing how inverses function in reflecting across these axes. Observing these symmetries not only helps in understanding the mathematics but also reinforces the significance of the \( y = x \) line as a critical feature of inverse relationships.
Other exercises in this chapter
Problem 62
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Evaluate the integrals. $$\int_{\pi / 6}^{\pi / 4} \frac{\csc ^{2} x d x}{1+(\cot x)^{2}}$$
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Use logarithmic differentiation to find the derivative of \(y\) with respect to the given independent variable. $$y=\frac{1}{t(t+1)(t+2)}$$
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