Problem 67 - Thomas Calculus in SI Units

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. $$y=e^{x}, \quad-3 \leq x \leq 5, \quad x_{0}=1$$

Step-by-Step Solution

Answer

The function is one-to-one due to a positive derivative. Inverse is \(y=\ln(x)\). Tangents found and plotted symmetrically.

Step 1: Plot the Function and Its Derivative

The function given is \(y=f(x) = e^x\). The derivative is \(f'(x) = e^x\). Plot \(e^x\) and \(e^x\) over the interval \(-3 \leq x \leq 5\). Since the derivative \(f'(x) = e^x > 0\) for all \(x\) in the interval, \(f\) is strictly increasing and thus one-to-one on the interval.

Step 2: Solve for Inverse Function

To find the inverse of \(f(x)=e^x\), solve \(y = e^x\) for \(x\). This yields \(x = \ln(y)\), so the inverse function is \(g(y) = \ln(y)\).

Step 3: Find Tangent Line to f at \(x_0 = 1\)

The slope of the tangent line at \(x_0 = 1\) is \(f'(1) = e\). The point of tangency is \((1, e^1) = (1, e)\). The equation of the tangent line is \(y - e = e(x - 1)\), or \(y = ex - e + e = ex\).

Step 4: Find Tangent Line to the Inverse Function

Theorem 1 states that if \(f\) is differentiable and \(f'(x_0) eq 0\), then \(g'(y_0) = \frac{1}{f'(g(y_0))}\). Here, \(y_0 = f(x_0) = e\). Thus, \(g'(e) = \frac{1}{f'(1)} = \frac{1}{e}\). So, the slope of the tangent line to \(g\) at \((e, 1)\) is \(\frac{1}{e}\). The tangent line equation is \(y - 1 = \frac{1}{e}(x - e)\).

Step 5: Plot and Discuss Symmetries

Plot \(f(x) = e^x\), \(g(y) = \ln(y)\), the identity function \(y = x\), the tangent line to \(f\) at \((1, e)\), and the tangent line to \(g\) at \((e, 1)\). Also, plot the line segment from \((1, e)\) to \((e, 1)\). Symmetrically, across the line \(y = x\), each point on \(f\) maps to a corresponding point on \(g\), illustrating a mirror image across \(y = x\).

Key Concepts

Derivatives of Inverse Functions

When dealing with the derivatives of inverse functions, a key point to remember is how they relate to one another. If you have a function \( f(x) \) and its inverse \( g(y) \), their derivatives are connected. According to Theorem 1 in calculus, if \( f \) is differentiable and \( f'(x_0) eq 0 \), then the derivative of the inverse function at a point is given by the formula:\[ g'(y_0) = \frac{1}{f'(g(y_0))} \]This tells us that the rate of change of the inverse function is the reciprocal of the rate of change of the original function at the corresponding point. In the context of the given exercise, where \( f(x) = e^x \) and \( g(y) = \ln(y) \), the derivative of \( g \), \( g'(e) = \frac{1}{f'(1)} = \frac{1}{e} \), shows this reciprocal nature.

Exponential and Logarithmic Functions

Exponential and logarithmic functions are intrinsically linked through their inverse relationship. An exponential function of the form \( y = e^x \) is a function that makes use of the constant \( e \) (approximately 2.718) as its base. This function grows rapidly as \( x \) increases due to its continuous rate of growth. On the other hand, the logarithmic function \( y = \ln(x) \) can be imagined as the "reverse" operation of an exponential operation.

• The exponential function \( e^x \) has no horizontal asymptotes, continually increasing over its domain.
• The function \( \ln(x) \), defined for \( x > 0 \), has a vertical asymptote at \( x = 0 \).

These two functions have a natural relationship where \( y = e^x \) and \( x = \ln(y) \), indicating their inverse properties.This inverse relationship is exemplified in the exercise by the functions \( f(x) = e^x \) and its inverse \( g(y) = \ln(y) \), which play a crucial role in various mathematical analyses.

Tangent Lines

Tangent lines serve as excellent linear approximators of functions at given points. When you draw a tangent line to a curve at a point, it touches the curve at just that point and has the same slope as the curve does at that location. This means that the slope of the tangent line is precisely the derivative of the function at that point.In the exercise given, at \( x_0 = 1 \) for the function \( f(x) = e^x \), the derivative \( f'(1) = e \) provides the slope of the tangent. The equation of the tangent line becomes \( y - e = e(x - 1) \), simplifying to \( y = ex \). This line shows how the function is changing at a point \( x = 1 \).Similarly, when considering its inverse \( g(y) = \ln(y) \), the tangent line at the symmetric point (\(e, 1\)) across the line \(y = x\) has a slope of \(\frac{1}{e}\), illustrating the principle of how derivatives of inverse functions are reciprocal. The tangent line here would be \( y - 1 = \frac{1}{e}(x - e) \).

One-to-One Functions

One-to-one functions have a unique property: every output value is produced by exactly one input value. This characteristic is crucial when we want to ensure that a function has an inverse because it guarantees that the inverse will be a well-defined function.For the function \( f(x) = e^x \) in the exercise, it is one-to-one over any interval because its derivative \( f'(x) = e^x \) is always positive. This means \( f(x) \) is strictly increasing across any interval, a clear indicator of its one-to-one nature.

• A function is one-to-one when for any two different inputs, the outputs do not overlap: if \( x_1 eq x_2, \) then \( f(x_1) eq f(x_2) \).
• Because \( e^x \) is continuously increasing, it will never repeat the same value for different inputs, thus ensuring it is one-to-one.

This property makes it possible for \( f(x) = e^x \) to have a well-defined inverse function \( g(y) = \ln(y) \), fulfilling the needs of various mathematical explorations, just like the one given in this exercise.