Problem 19 - Calculus Early Transcendentals

Question

a. Determine whether the Mean Value Theorem applies to the following functions on the given interval \([a, b]\). b. If so, find the point(s) that are guaranteed to exist by the Mean Value Theorem. c. For those cases in which the Mean Value Theorem applies, make a sketch of the function and the line that passes through \((a, f(a))\) and\((b, f(b)) .\) Mark the points \(P\) at which the slope of the function equals the slope of the secant line. Then sketch the tangent line at \(P\). $$f(x)=e^{x} ;[0, \ln 4]$$

Step-by-Step Solution

Answer

If so, what is the coordinate of that point? Answer: Yes, according to the Mean Value Theorem, there is a point on the function \(f(x) = e^x\) between \(x = 0\) and \(x = \ln 4\) where the slope of the tangent line equals the slope of the secant line. The coordinate of that point is \(\left(\ln{\left(\frac{3}{\ln 4}\right)}, \frac{3}{\ln 4}\right)\).

Step 1 - Checking continuity and differentiability

To apply the Mean Value Theorem, the function should be continuous on the closed interval and differentiable on the open interval. The function \(f(x) = e^x\) is continuous and differentiable everywhere, so it's continuous on the closed interval \([0, \ln 4]\) and differentiable on the open interval \((0, \ln 4)\).

Step 2 - Finding the Slope of the Secant Line

Now, we will find the slope of the secant line between \((a, f(a))\) and \((b, f(b))\). The slope, \(m_{sec}\), is given by: $$m_{sec} = \frac{f(b) - f(a)}{b - a}$$ Here, \(a = 0\) and \(b = \ln 4\). $$m_{sec} = \frac{f(\ln 4) - f(0)}{\ln 4 - 0}$$ Since \(f(x) = e^x\), we have \(f(0) = e^0 = 1\) and \(f(\ln 4) = e^{\ln 4} = 4\) $$m_{sec} = \frac{4 - 1}{\ln 4}$$ $$m_{sec} = \frac{3}{\ln 4}$$

Step 3 - Finding the points guaranteed by MVT

Now, we will find the point(s) \(c\) where the derivative of the function equals the slope of the secant line, i.e., \(f'(c) = m_{sec}\). The derivative of \(f(x) = e^x\) is also \(f'(x) = e^x\). So we are looking for \(c\) such that: $$e^c = \frac{3}{\ln 4}$$ Taking the natural logarithm of both sides, we get: $$c = \ln{\left(\frac{3}{\ln 4}\right)}$$ So, there's one point, \(P\), where the tangent line has the same slope as the secant line, at \(x = \ln{\left(\frac{3}{\ln 4}\right)}\) To find the coordinates of point \(P\), substitute the value of \(c\) into the function: $$f(c) = f\left(\ln{\left(\frac{3}{\ln 4}\right)}\right) = e^{\ln{\left(\frac{3}{\ln 4}\right)}} = \frac{3}{\ln 4}$$ So, \(P = \left(\ln{\left(\frac{3}{\ln 4}\right)}, \frac{3}{\ln 4}\right)\).

Step 4 - Sketching the function, secant line, and tangent line

Now, to summarize and make a sketch: 1. Sketch the function \(f(x) = e^x\). 2. Plot the end points of the interval \(A = (0, 1)\) and \(B = (\ln 4, 4)\). 3. Draw the secant line between points \(A\) and \(B\). 4. Locate the point \(P = \left(\ln{\left(\frac{3}{\ln 4}\right)}, \frac{3}{\ln 4}\right)\) on the graph. 5. Draw the tangent line at point \(P\) with the same slope as the secant line, which is \(\frac{3}{\ln 4}\). After completing the sketch, we will have the graph of \(f(x) = e^x\), along with the secant line between the points \((0,1)\) and \((\ln 4, 4)\), and also the tangent line at the point \(\left(\ln{\left(\frac{3}{\ln 4}\right)}, \frac{3}{\ln 4}\right)\) as guaranteed by the Mean Value Theorem.

Key Concepts

Continuity

The concept of continuity is crucial for applying the Mean Value Theorem (MVT). A function is considered continuous if there are no breaks, jumps, or holes in its graph within a specific interval. In mathematical terms, continuity at a point implies that the limit of the function as it approaches the point is equal to the function's value at that point.

For a function to be continuous on a closed interval \([a, b]\), it must also be continuous at the endpoints.

• This means the limits from both sides of the interval exist and match the function's value at those endpoints.
• For the given exercise, the exponential function \(f(x) = e^x\) is continuous on the interval \([0, \ln 4]\) because exponential functions are smooth and unbroken everywhere in their domain.

Understanding continuity ensures that we are examining a uniform graph without sudden or untraceable changes, which is fundamental for the sound application of mathematical theorems like the MVT.

Differentiability

Differentiability is another essential condition needed along with continuity for the application of the Mean Value Theorem. A function is said to be differentiable over an interval if it has a derivative at every point within that interval.

Differentiability implies that the function is smooth and doesn't have any sharp corners or cusps within the interval under consideration.

• On an open interval \((a, b)\), differentiability means the function possesses a defined tangent at each point.
• In our example, since the function \(f(x) = e^x\) is differentiable everywhere, this is satisfied over the interval \((0, \ln 4)\).

Understanding differentiability ensures that the slope of the function at different points can be calculated reliably, which allows us to find where this slope equals the slope of the secant line.

Exponential Functions

Exponential functions are a class of mathematical functions characterized by a constant base raised to a variable exponent, typically represented as \(e^x\), where \(e\) is Euler's number, approximately 2.718. These functions have unique properties that make them an excellent model for continuous growth processes.

Key characteristics of exponential functions include:

• Smooth and continuous curves where the rate of increase or decrease becomes progressively more rapid.
• The derivative of \(e^x\), denoted as \(f'(x)\), is \(e^x\) again, showing its unique self-replicating property.
• Always positive, moving through the y-axis at \(y = 1\) and continuously rising for \(x > 0\).

In the context of the given problem, these properties allow for easy calculation of changes over an interval (like finding the point where the function's slope equals the secant line's slope, as done with the MVT). The exponential's inherent properties ensure it's a model that provides useful applications in calculus and real-world scenarios.