Chapter 5: Problem 54
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 \(f(x)\).
Short Answer
Step by step solution
Show f(-2) = f(2)
Compute f'(x)
Show no c in (-2,2) makes f'(c)=0
Explain relation to Rolle's Theorem
Sketch the graph of f(x) using a graphing calculator
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Rolle's Theorem
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 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.
Absolute Value
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 \(x\).
Graphing Calculator
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.