Problem 55 - 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) \(f(-2)=f(2)\); (b) No \(c\) in \((-2,2)\) with \(f'(c)=0\); (d) Rolle's theorem not applicable.

Step 1: Evaluate function at boundary points

Given the function \(f(x) = e^{-|x|}\), we need to evaluate \(f(-2)\) and \(f(2)\). Since \(|-x| = x\), both \(f(-2)\) and \(f(2)\) are equal to \(e^{-2}\). Thus, \(f(-2) = f(2)\).

Step 2: Compute the derivative

To find \(f'(x)\), consider the absolute value function split: for \(x \geq 0\), \(f(x) = e^{-x}\), and for \(x < 0\), \(f(x) = e^{x}\). For \(x > 0\), \(f'(x) = -e^{-x}\). For \(x < 0\), \(f'(x) = e^{x}\). At \(x = 0\), \(f(x)\) is not differentiable due to the sharp vertex.

Step 3: Show there is no critical point in (-2, 2)

A critical point occurs where the derivative is zero. \(f'(x) = -e^{-x}\) for \(x > 0\) and \(f'(x) = e^{x}\) for \(x < 0\). Neither is zero within the interval \((-2, 0)\) and \((0, 2)\). Hence, there is no \(c\) such that \(f'(c) = 0\) in \((-2, 2)\).

Step 4: Explain Rolle's theorem applicability

Rolle's theorem requires \(f(x)\) to be continuous and differentiable on \((-2, 2)\). Though \(f(x)\) is continuous, it is not differentiable at \(x = 0\), as indicated by the sharp corner. Thus, Rolle's theorem conditions are not fully satisfied.

Step 5: Graph the function

Using a graphing calculator, sketch the graph of \(f(x) = e^{-|x|}\). The graph will show a peak at \(x = 0\) and decrease symmetrically towards \(x = -2\) and \(x = 2\), highlighting its symmetry and lack of critical points in \((-2, 2)\).

Key Concepts

Continuity

Continuity is a fundamental concept in calculus, which ensures that a function does not have any jumps, breaks, or holes over its domain. When a function is continuous on a closed interval, like \([-2, 2]\), it smoothly passes through every point between the interval's start and end. For the function \(f(x) = e^{-|x|}\), it is important to note that the exponential function itself is continuous over all real numbers. The only consideration here is the absolute value, which also remains continuous as it does not introduce abrupt changes in the graph.
The function \(f(x) = e^{-|x|}\) is therefore continuous over the interval \([-2, 2]\). This continuity means the graph of \(f(x)\) flows unbroken between these points. It is necessary for applying concepts like Rolle's Theorem, which requires continuity to even start the discussion. Always check continuity first when dealing with such exercises.

Differentiability

Differentiability asks whether a function can be differentiated, meaning whether the derivative exists at each point in its domain. For a function to be differentiable at a point, it must be smooth there without any sharp bends or cusps. Our function, \(f(x) = e^{-|x|}\), shifts in form across its domain:

• For \(x > 0\), we express it as \(f(x) = e^{-x}\).
• For \(x < 0\), it simplifies to \(f(x) = e^x\).

When you calculate its derivative, you'll see that:

• For \(x > 0\), \(f'(x) = -e^{-x}\).
• For \(x < 0\), \(f'(x) = e^x\).

However, at \(x = 0\), there is a sharp point, often referred to as a cusp, and hence it is not differentiable there. Differentiability is crucial because Rolle's Theorem requires the function to be differentiable on the open interval \((-2, 2)\). The lack of differentiability at \(x = 0\) is precisely why Rolle's Theorem doesn't apply directly here.

Critical Points

A critical point of a function is where the derivative is either zero or doesn't exist. These are important because they often correspond to peaks, troughs, or other points of interest on the graph of a function. To find critical points, you first find the derivative and then solve for points where this derivative is zero or undefined.
For \(f(x) = e^{-|x|}\), the derivative function, after analyzing the cases for \(x > 0\) and \(x < 0\), provides:

• For \(x > 0\), \( f'(x) = -e^{-x}\)
• For \(x < 0\), \( f'(x) = e^x\)

But notice that neither for any \(x > 0\) nor for any \(x < 0\) does the derivative equate to zero. Thus, there are no critical points for \(x\) within \((-2, 2)\), since the derivative never reaches zero.
The essence here is that despite the function having the same value at both edges of the interval, it is not turning or flattening out completely, a requirement by Rolle's Theorem.

Graphing Functions

Graphing a function is a powerful way to visually understand its behavior across an interval, especially to identify key features like symmetry, height, and critical points. For \(f(x) = e^{-|x|}\), sketching involves plotting points for different \(x\)-values and seeing how the curve shapes up.
Using a graphing calculator or software, we plot \(f(x)\) over the interval \([-2, 2]\). You will observe:

• The function is symmetric across the \(y\)-axis, due to the absolute value in its exponent.
• A peak or cusp is visible at \(x = 0\), consistent with our finding that it isn't differentiable there.
• No other peaks or troughs exist between \(-2\) and \(2\), confirming our analysis about the absence of critical points in the interval.

Seeing the function will make the theoretical concepts more intuitive, as you witness how the curve doesn't flatten out or acquire other critical characteristics within \((-2, 2)\). This visualization ties together the understanding of Rolle's Theorem, continuity, and differentiability.