Problem 40
Question
Find the critical numbers of \(f\) (if any). Find the open intervals on which the function is increasing or decreasing and locate all relative extrema. Use a graphing utility to confirm your results. $$ f(x)=(x-1) e^{x} $$
Step-by-Step Solution
Verified Answer
The function has a critical point and relative minimum at (0,-1). It is decreasing on the interval (-∞, 0) and increasing on the interval (0, ∞).
1Step 1: Finding the derivative of the function
Firstly, differentiate the given function \( f(x) = (x - 1)e^x \) using the product rule (question's derivative equals first function's derivative times second function plus first function times second function's derivative). The derivative function \( f'(x) \) is as follows: \( f'(x) = e^x(x - 1) + e^x \) \nSimplify this to:\( f'(x) = x e^x \)
2Step 2: Finding the critical points
Set \( f'(x) \) equal to zero and solve for \( x \) to find the critical points.\( x e^x = 0\)From this we can see that the critical points are \(x = 0\)
3Step 3: Determine Intervals of Increase and Decrease
Divide the number line into intervals using the critical point \(x = 0\) as a boundary, then choose a test point from each interval and substitute it into the first derivative.Interval (-∞, 0), test point: \( x = -1\). Substituting \(x = -1\) into the derivative gives a negative number.This indicates the function is decreasing on this interval (-∞, 0).Interval (0, ∞), test point: \( x = 1\). Substituting \(x = 1\) into the derivative gives a positive number.This indicates the function is increasing on this interval (0, ∞). So, the function decreases on (-∞, 0) and increases on (0, ∞).
4Step 4: Determine the local maximum and minimum
A change from decreasing to increasing at a point signifies a local (relative) minimum, and a change from increasing to decreasing is a local (relative) maximum. In this case, f(x) changes from decreasing to increasing at x=0, so the function has a relative minimum at the point (0, f(0)) or (0,-1).
5Step 5: Confirm with graph
Using a graphing utility, plot the function \( f(x) = (x - 1) e^x \). You should see that it decreases to the point (0,-1), then starts increasing, which validates your previous findings.
Key Concepts
Product RuleDerivativeIncreasing and Decreasing IntervalsRelative Extrema
Product Rule
The product rule is an essential derivative rule used when taking the derivative of a function that is the product of two or more functions. If we have a function \( h(x) = f(x) \) times \( g(x) \), where both \( f(x) \) and \( g(x) \) are differentiable, the product rule tells us that the derivative of \( h(x) \) is \( h'(x) = f'(x)g(x) + f(x)g'(x) \). This allows us to differentiate functions that would be difficult to manipulate using algebra alone.
For instance, in our exercise where \( f(x) = (x - 1)e^x \), \( f(x) \) is the product of \( (x - 1) \) and \( e^x \). The derivative \( f'(x) \) is found using the product rule, resulting in \( f'(x) = e^x(x - 1) + e^x \). It's one of the critical steps in determining the function's behavior and finding critical numbers.
For instance, in our exercise where \( f(x) = (x - 1)e^x \), \( f(x) \) is the product of \( (x - 1) \) and \( e^x \). The derivative \( f'(x) \) is found using the product rule, resulting in \( f'(x) = e^x(x - 1) + e^x \). It's one of the critical steps in determining the function's behavior and finding critical numbers.
Derivative
The derivative of a function at a certain point is a measure of how much the function's output value changes in response to a change in the input value near that point. In simpler terms, it represents the slope of the tangent line to the function's graph at that point, or the rate of change. Calculating the derivative is a cornerstone of calculus and is essential in finding critical points where the function's rate of change is zero.
The derivative of \( f(x) \) in our example is \( f'(x) = xe^x \). Setting this equal to zero gives us the function's critical numbers, which are invaluable for analyzing the function's increasing and decreasing intervals and locating relative extrema.
The derivative of \( f(x) \) in our example is \( f'(x) = xe^x \). Setting this equal to zero gives us the function's critical numbers, which are invaluable for analyzing the function's increasing and decreasing intervals and locating relative extrema.
Increasing and Decreasing Intervals
Understanding increasing and decreasing intervals of a function is vital for analyzing its overall behavior. A function is said to be increasing on an interval if, as \( x \) moves from left to right, \( f(x) \) goes up. Conversely, a function is decreasing on an interval if \( f(x) \) goes down as \( x \) moves from left to right. Critical numbers can help partition the domain into intervals where we can test the sign of the derivative to determine whether the function is increasing or decreasing.
In the example \( f(x) = (x - 1) e^x \), the critical number \( x = 0 \) is used to divide the number line into two intervals. We test points from each interval in the derivative to ascertain the function's behavior: decreasing on \( (-\textbackslash infty , 0) \) and increasing on \( (0, \textbackslash infty) \) which leads us to identify the nature of the extremum at the critical number.
In the example \( f(x) = (x - 1) e^x \), the critical number \( x = 0 \) is used to divide the number line into two intervals. We test points from each interval in the derivative to ascertain the function's behavior: decreasing on \( (-\textbackslash infty , 0) \) and increasing on \( (0, \textbackslash infty) \) which leads us to identify the nature of the extremum at the critical number.
Relative Extrema
Relative extrema, often referred to as local maxima or minima, are points on the graph of a function where the function value is at its highest or lowest respectively compared to its immediate surroundings. To find these points, we look for changes in the function's increasing or decreasing behavior. If the function switches from decreasing to increasing at a certain point, that point is a relative minimum. Conversely, a switch from increasing to decreasing indicates a relative maximum.
In our exercise, by finding that the function changes from decreasing to increasing at the critical number \( x = 0 \), we are able to conclude that there is a relative minimum at the point \( (0, f(0)) \) or \( (0, -1) \) by evaluating the function at \( x = 0 \). These points are important in understanding the full picture of the function's behavior and can be visually confirmed using a graphing utility.
In our exercise, by finding that the function changes from decreasing to increasing at the critical number \( x = 0 \), we are able to conclude that there is a relative minimum at the point \( (0, f(0)) \) or \( (0, -1) \) by evaluating the function at \( x = 0 \). These points are important in understanding the full picture of the function's behavior and can be visually confirmed using a graphing utility.
Other exercises in this chapter
Problem 40
Use a graphing utility to (a) graph the function \(f\) on the given interval, (b) find and graph the secant line through points on the graph of \(f\) at the end
View solution Problem 40
In Exercises \(37-40\), use a graphing utility to graph the function and identify any horizontal asymptotes. $$ f(x)=\frac{\sqrt{9 x^{2}-2}}{2 x+1} $$
View solution Problem 41
The range \(R\) of a projectile fired with an initial velocity \(v_{0}\) at an angle \(\theta\) with the horizontal is \(R=\frac{v_{0}^{2} \sin 2 \theta}{g},\)
View solution Problem 41
In Exercises \(41-44,\) use a computer algebra system to analyze the function over the given interval. (a) Find the first and second derivatives of the function
View solution