Question

In Exercises \(11-18,\) use analytic methods to find the extreme values of the function on the interval and where they occur. $$g(x)=e^{-x}, \quad-1 \leq x \leq 1$$

Short answer

The maximum value of the function is \(e\) at \(x=-1\), and the minimum value is \(1/e\) at \(x=1\).

Step-by-step solution

Find the first derivative

The first derivative of the function \(g(x) = e^{-x}\) is found using the chain rule for differentiation, which gives \(g'(x)=-e^{-x}\). This derivative is neither zero nor undefined for any real x.

Identify the critical points

As there are no x-values for which \(g'(x)\) is zero or undefined, the critical points must occur at the endpoints of the given interval. These are -1 and 1.

Determine the function values at the critical points

The function values at the critical points are obtained by substituting the values -1 and 1 into the original function to get \(g(-1)=e^{1}=e\) and \(g(1)=e^{-1}=1/e\). These are the extreme values of the function within the given interval. Since \(e>1\), we have \(e\) as the maximum value and \(1/e\) as the minimum.

Identify where the extreme values occur

The maximum value \(e\) occurs at \(x=-1\) and the minimum value \(1/e\) occurs at \(x=1\).

Key concepts

First Derivative Test

The First Derivative Test is a crucial procedure in calculus used to determine where a function has relative maximum or minimum points. To implement the test, we first identify the critical points of a function by finding where its first derivative is zero or undefined. After identifying these points, we examine the behavior of the first derivative on either side of these points.

When applying the First Derivative Test to the function at hand, we take into account the interval \( -1 \leq x \leq 1 \). However, in this particular exercise, \( g'(x) \) does not equal zero or become undefined within the interval, prompting us to look at the interval's endpoints to find the extreme values. The First Derivative Test here leads us to conclude that no relative extrema exist inside the interval, and we must evaluate only the endpoints. To enhance understanding, remember that a change in sign of the first derivative moving through a critical point suggests a relative extremum at that point.

Critical Points in Calculus

Critical points in calculus are locations on the graph of a function where the function's derivative is either zero or undefined. These points are valuable because they can potentially be the locations of relative maximums, relative minimums, or points of inflection. It is important to note that not all critical points are locations of relative extrema; some may be points where the function continues without a peak or a trough.

In the given exercise, since the derivative \( g'(x) = -e^{-x} \) is never zero or undefined, traditional critical points do not exist within the open interval \( -1 < x < 1 \). However, the endpoints of the interval themselves can sometimes act as critical points when considering extreme values on a closed interval. In this case, \( x = -1 \) and \( x = 1 \) function as critical points for our analysis, which is why we need to examine the function's value at these points to identify the absolute extrema.

Chain Rule for Differentiation

The chain rule is a formula used to compute the derivative of a composite function. If a function y is composed of a function u which is in turn a function of x, i.e., \( y = f(u) \) and \( u = g(x) \) then the derivative of y with respect to x is the product of the derivative of y with respect to u and the derivative of u with respect to x. This is succinctly written as \( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \).

In the context of the exercise, the function \( g(x) = e^{-x} \) appears simple, but it's actually a composite function where \( u = -x \) and \( y = e^u \). Using the chain rule, we differentiate \( e^u \) with respect to u to get \( e^u \) and then differentiate \( u \) with respect to x to get -1, ultimately multiplying them together to yield \( g'(x) = -e^{-x} \). This demonstration of the chain rule in action serves as a fundamental example of how derivatives are taken for composite functions, and understanding it is pivotal for calculus students.