Problem 54 - Calculus for Biology and Medicine

Question

Suppose that \(f(x)=e^{-|x|}, x \in[-2,2]\). (a) Show that \(f(-2)=f(2)\). (b) Compute \(f^{\prime}(x)\), where defined. (c) Show that there is no number \(c \in(-2,2)\) such that \(f^{\prime}(c)=0\). (d) Explain why your results in (a) and (c) do not contradict Rolle's theorem. (e) Use a graphing calculator to sketch the graph of \(f(x)\).

Step-by-Step Solution

Answer

a: Yes, f(-2)=f(2). b: f'(x) is not zero for any x in (-2,2). c: No such c exists. d: f is not fully differentiable as Rolle's theorem requires.

Step 1: Show f(-2) = f(2)

To show that \(f(-2) = f(2)\), we simply evaluate the function at \(x = -2\) and \(x = 2\). The function is given by \(f(x) = e^{-|x|}\). So, \(f(-2) = e^{-|-2|} = e^{-2}\) and \(f(2) = e^{-|2|} = e^{-2}\). Thus, \(f(-2) = f(2)\).

Step 2: Compute f'(x)

To find \(f'(x)\), we need to consider \(f(x) = e^{-|x|}\) split into two cases based on the definition of absolute value: for \(x \geq 0\), \(|x| = x\) and for \(x < 0\), \(|x| = -x\). 1. For \(x > 0\), \(f(x) = e^{-x}\), so \(f'(x) = -e^{-x}\).2. For \(x < 0\), \(f(x) = e^{x}\), so \(f'(x) = e^{x}\). 3. At \(x = 0\), we find that \(f\) is not differentiable because the left-hand and right-hand derivatives at 0 do not match.

Step 3: Show no c in (-2,2) makes f'(c)=0

To check if there exists \(c \in (-2, 2)\) such that \(f'(c) = 0\), use the results from Step 2:- For \(x > 0\), \(f'(x) = -e^{-x}\), which is never 0.- For \(x < 0\), \(f'(x) = e^{x}\), which is also never 0.Since \(f\) is not differentiable at \(x = 0\) and in both regions of the domain \(f'(x)\) is never zero, there is no \(c\) such that \(f'(c) = 0\).

Step 4: Explain relation to Rolle's Theorem

Rolle's theorem states that if \(f\) is continuous on \([a, b]\) and differentiable on \((a, b)\) and \(f(a) = f(b)\), then there exists at least one \(c \in (a, b)\) such that \(f'(c) = 0\). In this case, while \(f(-2) = f(2)\), the function \(f(x) = e^{-|x|}\) is not differentiable at \(x = 0\). Thus, Rolle's theorem does not apply because \(f\) is not differentiable on the entire interval \((-2, 2)\).

Step 5: Sketch the graph of f(x) using a graphing calculator

To visually understand \(f(x)\), input \(f(x) = e^{-|x|}\) into a graphing calculator for \(x \in [-2, 2]\). You should see that the graph is symmetric about the y-axis with a peak at \(x = 0\) and decreasing as \(|x|\) increases. This confirms that \(f(-2) = f(2)\), and visually supports the earlier analyses of function symmetry and differentiability.

Key Concepts

Rolle's Theorem

Rolle's Theorem is a fascinating concept in calculus that connects the dots between the values of a function at two points and its slope in between. This theorem states that if a function, say \(f(x)\), is continuous on the closed interval \([a, b]\) and differentiable on the open interval \((a, b)\), with \(f(a) = f(b)\), then there must be at least one point \(c\) in the interval \((a, b)\) where the derivative \(f'(c) = 0\). In simpler terms, there exists a point where the tangent to the curve is horizontal.

In the exercise, we're dealing with \(f(x) = e^{-|x|}\), and we found that \(f(-2) = f(2)\). However, since the function is not differentiable at \(x = 0\), Rolle's Theorem does not apply. Differentiability is a key condition here that ensures the existence of such a \(c\). When this condition is not met throughout the interval, the theorem cannot be used.

Derivative

The derivative of a function provides critical information about the rate at which the function is changing. It's essentially the slope of the tangent to the function at any given point. For the function \(f(x) = e^{-|x|}\), we computed the derivative for different portions of the domain.

The function behaves differently based on the value of \(x\):

• For \(x \geq 0\), the absolute value \(|x|\) equals \(x\), and therefore, \(f(x) = e^{-x}\). In this case, the derivative \(f'(x) = -e^{-x}\).
• For \(x < 0\), the absolute value \(|x|\) equals \(-x\), turning the function into \(f(x) = e^{x}\), leading to \(f'(x) = e^{x}\).
• At \(x = 0\), however, \(f(x)\) is not differentiable because the left-hand and right-hand derivatives do not match. This causes a break in the curve's smoothness that prevents calculating a single slope.

Understanding this behavior helps uncover why there's no \(c\) in \((-2, 2)\) where \(f'(c) = 0\).

Absolute Value

Absolute values often appear in mathematical expressions to indicate a number's distance from zero, stripping any sign it may have. In the function \(f(x) = e^{-|x|}\), it involves the exponential function and the absolute value in the exponent. This combination affects the function's general shape and symmetry.

Notice how the absolute value \(|x|\) factors into defining two distinct parts for \(f(x)\):

• When \(x \geq 0\), it directly means \(|x| = x\), leading the expression in the exponential to become \(-x\).
• When \(x < 0\), it turns into \(-x\), simplifying the expression inside the exponential to \(x\).

Through these transformations, we can observe the symmetry around the y-axis. This symmetry is key to why \(f(-2)\) and \(f(2)\) are the same. While absolute values make calculations more involved, they add unique characteristics to a function's behavior.

Graphing Calculator

A graphing calculator is a powerful tool that visually represents complex functions and their behaviors. It's especially useful when words and numbers fall short in conveying a function's full story. For example, inputting \(f(x) = e^{-|x|}\) into a graphing calculator allows you to see its symmetry and how the graph sharpens and declines away from \(x = 0\).

Here's what to expect when using a graphing calculator for this exercise:

• The graph is symmetric about the y-axis, illustrating the concept of absolute values.
• It has a peak at \(x = 0\), showing how the function reaches its maximum value there.
• As \(|x|\) increases, the function decreases rapidly, which aligns with the computed derivatives.

This visual support strengthens understanding and provides a vivid context for theoretical concepts like symmetry, differentiability, and the impact of absolute values.