Problem 64
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. $$y=\frac{x^{3}}{x^{2}+1}, \quad-1 \leq x \leq 1, \quad x_{0}=1 / 2$$
Step-by-Step Solution
Verified Answer
Function \( f \) is one-to-one on \([-1, 1]\), the tangent lines and functions are symmetric about \( y = x \).
1Step 1: Plot the Function and its Derivative
For the function \( y = f(x) = \frac{x^3}{x^2 + 1} \), we first find its derivative. Using the quotient rule, find \[f'(x) = \frac{(3x^2)(x^2 + 1) - (x^3)(2x)}{(x^2 + 1)^2} = \frac{3x^4 + 3x^2 - 2x^4}{(x^2 + 1)^2} = \frac{x^4 + 3x^2}{(x^2 + 1)^2}\]Plotting this function and its derivative over the interval \(-1 \leq x \leq 1\), shows that the derivative is non-negative, except at isolated points. Since the derivative is non-negative and zero at only a single point, \( f \) is one-to-one over the given interval.
2Step 2: Find the Inverse Function
To find the inverse, solve \( y = \frac{x^3}{x^2 + 1} \) for \( x \). This requires algebraic manipulation. Assume that the inverse function is \( g(y) = x \), and solve for \( x \):\[y(x^2 + 1) = x^3\]This equation can be simplified to:\[x^3 - yx^2 - y = 0\]The inverse function can be found implicitly, but it may involve numerical methods to find specific values.
3Step 3: Find the Tangent Line to f at a Specific Point
The point is \( \left( x_0, f(x_0) \right) \) with \( x_0 = \frac{1}{2} \). Calculate \( f\left( \frac{1}{2} \right) = \frac{\left( \frac{1}{2} \right)^3}{\left( \frac{1}{2} \right)^2 + 1} = \frac{1/8}{5/4} = \frac{1}{10} \). Find the derivative at this point: \[f'(\frac{1}{2}) = \frac{(\frac{1}{2})^4 + 3(\frac{1}{2})^2}{(\frac{1}{2})^2 + 1)^2} = \frac{1/16 + 3/4}{25/16} = \frac{13/16}{25/16} = \frac{13}{25}\]Thus, the equation of the tangent line is \[y - \frac{1}{10} = \frac{13}{25}(x - \frac{1}{2})\]
4Step 4: Find the Tangent Line to the Inverse Function
The point is \( \left( f(\frac{1}{2}), \frac{1}{2} \right) = \left( \frac{1}{10}, \frac{1}{2} \right) \). By Theorem 1, the slope of the tangent line at this point is the reciprocal of the derivative at \( x_0 = \frac{1}{2} \), giving the slope \( \frac{25}{13} \).The equation of the tangent line is \[y - \frac{1}{2} = \frac{25}{13}(x - \frac{1}{10})\]
5Step 5: Plot All Components and Discuss Symmetries
Plot the function \( f \) and its inverse \( g \), the identity line \( y = x \), and the tangent lines derived in previous steps. Also include the line segment joining the two key points \( \left( \frac{1}{2}, \frac{1}{10} \right) \) and \( \left( \frac{1}{10}, \frac{1}{2} \right) \). Notice that the graph of \( f \) and \( g \), along with the tangent lines, show a symmetry about the line \( y = x \), with each tangent line being a mirror image about this line.
Key Concepts
Tangent LineOne-to-One FunctionDerivativeSymmetry
Tangent Line
A tangent line is a straight line that just "touches" a curve at a single point, known as the point of tangency. At this point, the tangent line has the same slope as the curve it touches. In practical terms, the tangent line can be thought of as the line that locally approximates the curve at a given point. Understanding tangent lines is vital as they help in approximating the behavior of the curve and analyze the rate of change.
To find the equation of a tangent line to a function at a particular point, you must know two things: the slope of the tangent line and the coordinates of the point where it touches the curve. The slope is found by calculating the derivative of the function at that point.
For example, in the function given in the exercise, the derivative was used to find the slope at the specified point, and with the point's coordinates, the equation of the tangent line was derived. By using the formula for a straight line, you can express the tangent line as:
To find the equation of a tangent line to a function at a particular point, you must know two things: the slope of the tangent line and the coordinates of the point where it touches the curve. The slope is found by calculating the derivative of the function at that point.
For example, in the function given in the exercise, the derivative was used to find the slope at the specified point, and with the point's coordinates, the equation of the tangent line was derived. By using the formula for a straight line, you can express the tangent line as:
- If the function is \(f(x)\) and the point of tangency is \((x_0, f(x_0))\), then the tangent line equation is:
\[y - f(x_0) = f'(x_0)(x - x_0)\]
One-to-One Function
A one-to-one function is one in which every element of the range corresponds to exactly one element of the domain. This special property ensures that the function also has a well-defined inverse because no two different \(x\) values will have the same \(y\) value. This is crucial in determining whether a function is invertible.
To check if a function is one-to-one over a particular interval, you can look at its derivative. If the derivative is always positive or always negative within that interval, the function is strictly increasing or decreasing, respectively, indicating it's one-to-one. However, if the derivative is zero at any significant range, the function may not be one-to-one over that region.
In the context of the previous exercise, the derivative of the function was found to be non-negative across the interval \([-1, 1]\), only reaching zero at isolated points. This allowed the conclusion that the function is one-to-one over this interval. When dealing with such functions, knowing they are one-to-one provides confidence that you can meaningfully compute its inverse.
To check if a function is one-to-one over a particular interval, you can look at its derivative. If the derivative is always positive or always negative within that interval, the function is strictly increasing or decreasing, respectively, indicating it's one-to-one. However, if the derivative is zero at any significant range, the function may not be one-to-one over that region.
In the context of the previous exercise, the derivative of the function was found to be non-negative across the interval \([-1, 1]\), only reaching zero at isolated points. This allowed the conclusion that the function is one-to-one over this interval. When dealing with such functions, knowing they are one-to-one provides confidence that you can meaningfully compute its inverse.
Derivative
Derivatives represent the rate of change of a function with respect to a variable and are foundational to calculus. When considering the graph of a function, the derivative at any given point provides the slope of the tangent line at that point, which can visually be interpreted as how steep the function is at that juncture.
Calculating a derivative involves using rules like the power rule, product rule, quotient rule, or chain rule, depending on the function's structure. In the given exercise, the function was a quotient of two expressions, requiring the use of the quotient rule:
Calculating a derivative involves using rules like the power rule, product rule, quotient rule, or chain rule, depending on the function's structure. In the given exercise, the function was a quotient of two expressions, requiring the use of the quotient rule:
- Quotient Rule: For a function \(\frac{u(x)}{v(x)}\), the derivative \((\frac{d}{dx}[\frac{u}{v}])\) is given by:
\[ \frac{v(x)u'(x) - u(x)v'(x)}{(v(x))^2} \]
Symmetry
In mathematics, symmetry refers to a balanced and proportionate similarity found in two halves of an object. For functions, symmetry can be observed with respect to certain lines or points. Understanding symmetry is quite helpful in analyzing and simplifying problems involving functions and their graphs.
In the specific exercise, symmetry comes into play in relation to the inverse functions. The original function \(f\) and its inverse \(g\) displayed symmetry around the line \(y = x\), which represents the identity function. This diagonal line serves as a mirror, reflecting points from \(f\) onto their corresponding inverse points in \(g\) and vice versa.
Such symmetry can also be observed in the corresponding tangent lines at points across the line \(y = x\). Each tangent line to \(f\) at a point \( (x_0, f(x_0)) \) is mirrored by a tangent line to \(g\) at the point \((f(x_0), x_0)\), showing that the slopes of these lines are reciprocal, giving each line symmetric characteristics.
In the specific exercise, symmetry comes into play in relation to the inverse functions. The original function \(f\) and its inverse \(g\) displayed symmetry around the line \(y = x\), which represents the identity function. This diagonal line serves as a mirror, reflecting points from \(f\) onto their corresponding inverse points in \(g\) and vice versa.
Such symmetry can also be observed in the corresponding tangent lines at points across the line \(y = x\). Each tangent line to \(f\) at a point \( (x_0, f(x_0)) \) is mirrored by a tangent line to \(g\) at the point \((f(x_0), x_0)\), showing that the slopes of these lines are reciprocal, giving each line symmetric characteristics.
Other exercises in this chapter
Problem 64
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Use logarithmic differentiation to find the derivative of \(y\) with respect to the given independent variable. $$y=\frac{\theta \sin \theta}{\sqrt{\sec \theta}
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