Problem 20 - 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 \(P\). $$f(x)=\ln 2 x ;[1, e]$$

Step-by-Step Solution

Answer

If so, find the point(s) guaranteed to exist by the theorem and make a sketch. Yes, the Mean Value Theorem (MVT) applies to the function f(x) = ln(2x) on the interval [1, e], since the function is continuous and differentiable on this interval. By applying the MVT, we find that the point guaranteed to exist by the theorem is P = (e-1, ln(2(e-1))). The function, secant line, and tangent line can be sketched accordingly.

Step 1: Understand the Mean Value Theorem

The Mean Value Theorem states that if a function is continuous on a closed interval \([a, b]\) and differentiable on the open interval \((a, b)\), then there exists at least one point \(c\) in the interval \((a, b)\) such that $$f'(c) = \frac{f(b) - f(a)}{b - a}$$

Step 2: Determine the continuity and differentiability of the given function

The given function is \(f(x) = \ln 2x\). This function is continuous and differentiable for all \(x > 0\), since the natural logarithm is continuous and differentiable for all positive numbers. The interval specified is \([1, e]\), and since \(1\) and \(e\) are greater than 0, the function is continuous and differentiable on this interval.

Step 3: Apply the Mean Value Theorem

Since the function is continuous and differentiable on \([1, e]\), the Mean Value Theorem applies. We will now compute the derivative of the function and then find the point \(c\) where \(f'(c) = \frac{f(e) - f(1)}{e-1}\).

Step 4: Compute the derivative of the function

To find the derivative of \(f(x) = \ln (2x)\), we apply the chain rule: $$f'(x) = \frac{d(\ln (2x))}{dx}=\frac{1}{2x} \cdot \frac{d(2x)}{dx}=\frac{1}{2x} \cdot 2 =\frac{1}{x}$$

Step 5: Compute the value of the slope

Now we need to compute the slope of the secant line, which is given by: $$\frac{f(e)-f(1)}{e-1}=\frac{\ln(2e)-\ln(2)}{e-1}=\frac{\ln 2+\ln e-\ln 2}{e-1}=\frac{\ln e}{e - 1}=\frac{1}{e-1}$$

Step 6: Find the point c where f'(c) equals the slope of the secant line

We will now equate the derivative of the function, \(f'(x)\), to the slope computed in step 5, and solve for \(x\) to find the point \(c\): $$\frac{1}{x} = \frac{1}{e-1}$$ Multiplying both sides by \(x(e-1)\), we get $$x=e-1$$ So, \(c=e-1\).

Step 7: Find the point P

The point \(P\) is \((c, f(c))=(e-1, \ln(2(e-1)))\)

Step 8: Sketch the function, secant line, and tangent line

We can now sketch the function, \(f(x) = \ln(2x)\), on the interval \([1, e]\). Plot the points \((1, f(1))\) and \((e, f(e))\), and draw the secant line connecting these points. Also, mark the point \(P\) on the graph at \((e-1, \ln(2(e-1)))\), and draw the tangent line at this point with a slope equal to the slope of the secant line. In summary, the Mean Value Theorem applies to the function \(f(x) = \ln 2x\) on the interval \([1, e]\), and the point guaranteed to exist by the theorem is \(P=(e-1, \ln(2(e-1)))\). The function, secant line, and tangent line can be sketched accordingly.

Key Concepts

Continuity

Continuity is a crucial concept when working with the Mean Value Theorem. A function is continuous over an interval if there are no breaks, jumps, or holes in the graph on that interval. For the function \(f(x) = \ln(2x)\), we need it to be continuous on the interval \([1, e]\). The natural logarithm \(\ln(x)\) is continuous for all positive \(x\), and since \(2x\) is positive on \([1, e]\), we can conclude that \(\ln(2x)\) is continuous as well.
The importance of continuity lies in the Mean Value Theorem itself. If a function isn't continuous over the interval \([a, b]\), the theorem can't be applied.

• Check if the function is defined everywhere in \([a, b]\).
• Ensure there are no disruptions in the graph within the interval.

Understanding how to determine continuity will help in confidently applying the Mean Value Theorem, paving the way to more complex calculus concepts.

Differentiability

Differentiability means that a function has a derivative at all points in the interval you’re looking at, except possibly at the endpoints. For the Mean Value Theorem to apply to \(f(x) = \ln(2x)\), it must be differentiable on \((1, e)\). As we know, the natural logarithm is differentiable for all positive values of \(x\), so is \(\ln(2x)\).
To understand differentiability, keep these points in mind:

• The derivative of \(f(x) = \ln(2x)\) is \(f'(x) = \frac{1}{x}\).
• This derivative exists for all \(x > 0\) meaning \(\ln(2x)\) is differentiable as required.

Differentiability ensures a smooth curve without sharp corners or vertical tangents.
When combined with continuity, differentiability allows for the seamless application of critical calculus concepts like the Mean Value Theorem.

Natural Logarithm

The natural logarithm, denoted as \(\ln(x)\), is the logarithm to the base \(e\), where \(e\) is approximately 2.718. This function is only defined for positive \(x\). In calculus, the natural logarithm is important because it naturally relates to constant growth processes. For the function \(f(x) = \ln(2x)\), we're specifically interested in how it behaves from \(x = 1\) to \(x = e\).
Here are some key points about \(\ln(x)\):

• \(\ln(x)\) is continuous and smooth for \(x > 0\).
• The derivative \(\frac{d}{dx}[\ln(x)] = \frac{1}{x}\) highlights its rate of change.

By understanding \(\ln(x)\), one can grasp more advanced applications, like logarithmic differentiation. This knowledge helps in solving real-world problems involving growth and decay. When dealing with \(\ln(2x)\), knowing these basics allows you to efficiently apply the Mean Value Theorem to uncover key points like \(c = e-1\), deepening your comprehension of calculus functions.