Problem 23
Question
Find the absolute maximum and minimum values of each function on the given interval. Then graph the function. Identify the points on the graph where the absolute extrema occur, and include their coordinates. $$f(x)=x^{2}-1, \quad-1 \leq x \leq 2$$
Step-by-Step Solution
Verified Answer
Absolute max: 3 at \(x = 2\); Absolute min: -1 at \(x = 0\).
1Step 1: Identify critical points
First, find the derivative of the function \(f(x) = x^2 - 1\). The derivative is \(f'(x) = 2x\). Set the derivative equal to zero to find critical points: \(2x = 0\), which gives \(x = 0\). Check if this critical point is within the interval \([-1, 2]\). It is.
2Step 2: Evaluate function at critical points
Now calculate \(f(x)\) at the critical point \(x = 0\). Substitute \(x = 0\) into the original function: \(f(0) = 0^2 - 1 = -1\).
3Step 3: Evaluate function at interval endpoints
Evaluate \(f(x)\) at the endpoints of the interval. For \(x = -1\), \(f(-1) = (-1)^2 - 1 = 0\). For \(x = 2\), \(f(2) = 2^2 - 1 = 3\).
4Step 4: Determine absolute extrema
Compare the function values obtained: \(f(-1) = 0\), \(f(0) = -1\), and \(f(2) = 3\). The absolute maximum value is 3 at \(x = 2\), and the absolute minimum value is -1 at \(x = 0\).
Key Concepts
DerivativeCritical PointAbsolute MaximumAbsolute Minimum
Derivative
The derivative of a function is a fundamental concept in calculus. It's all about determining the rate at which the function's value changes as its input changes. In simpler terms, the derivative tells us how steep the function is or how its slope behaves at any point. If you imagine a curve on a graph, the derivative gives you a straight line, or the tangent, that just "kisses" the curve at a particular point without cutting through it.
To find the derivative of a function like \[ f(x) = x^2 - 1 \] we use differentiation rules. For our function, the derivative is\[ f'(x) = 2x, \] which means at any point \( x \), the slope of the tangent to the curve is \( 2x \). Here are quick tips to keep in mind:
To find the derivative of a function like \[ f(x) = x^2 - 1 \] we use differentiation rules. For our function, the derivative is\[ f'(x) = 2x, \] which means at any point \( x \), the slope of the tangent to the curve is \( 2x \). Here are quick tips to keep in mind:
- Derivatives indicate increasing or decreasing behavior.
- A positive derivative means the function is increasing.
- A negative derivative means the function is decreasing.
Critical Point
Critical points are very important when analyzing a function. They occur where the derivative equals zero or doesn't exist. This helps us identify where the function reaches a plateau, meaning it neither increases nor decreases just at that point.
For example, with the function \( f(x) = x^2 - 1 \), we determined that the derivative is \( f'(x) = 2x \). Setting the derivative equal to zero gives us: \[ 2x = 0 \] which simplifies to \( x = 0 \). This is our critical point.
To decide whether this point is a maximum, minimum, or neither, we must evaluate the function at this value and consider the interval endpoints as well.
For example, with the function \( f(x) = x^2 - 1 \), we determined that the derivative is \( f'(x) = 2x \). Setting the derivative equal to zero gives us: \[ 2x = 0 \] which simplifies to \( x = 0 \). This is our critical point.
To decide whether this point is a maximum, minimum, or neither, we must evaluate the function at this value and consider the interval endpoints as well.
Absolute Maximum
The absolute maximum is the highest point a function reaches over a given interval. It’s crucial in situations where you need to know the highest threshold of an outcome or data, like finding the highest point of a graph for a practical task.
In our example, to identify this, we find the function values at the critical points and at the interval endpoints. Compare:
The function clearly hits its peak at \( x = 2 \) with a value of 3, making this the absolute maximum. Remember:
In our example, to identify this, we find the function values at the critical points and at the interval endpoints. Compare:
- \(f(-1) = 0\)
- \(f(0) = -1\)
- \(f(2) = 3\)
The function clearly hits its peak at \( x = 2 \) with a value of 3, making this the absolute maximum. Remember:
- Check both critical points and endpoints.
- Look for where the function value is the greatest.
- The absolute maximum gives you a clear boundary of the function's growth in that interval.
Absolute Minimum
The absolute minimum is the lowest point that the function reaches over a given interval, providing insight into the floor or base level of the outcomes of a dataset or real-world scenario.
In our situation:
In our situation:
- We compare the function's values at its critical point and the endpoints: '
- \(f(-1) = 0\)
- \(f(0) = -1\)
- \(f(2) = 3\)
- Evaluate at critical points and endpoints.
- Identify where the function value is at its lowest.
- The absolute minimum is key to understanding the lowest bounds of behavior in a function's interval.
Other exercises in this chapter
Problem 22
Use I'Hópital's rule to find the limits. $$\lim _{x \rightarrow x / 2} \frac{\ln (\csc x)}{(x-(\pi / 2))^{2}}$$
View solution Problem 22
The graphs of \(y=\sqrt{x}\) and \(y=3-x^{2}\) intersect at one point \(x=r .\) Use Newton's method to estimate the value of \(r\) to four decimal places.
View solution Problem 23
Identify the coordinates of any local and absolute extreme points and inflection points. Graph the function. $$y=x+\sin x, \quad 0 \leq x \leq 2 \pi$$
View solution Problem 23
Find an antiderivative for each function. Do as many as you can mentally. Check your answers by differentiation. a. \(\frac{2}{\sqrt{1-x^{2}}}\) b. \(\frac{1}{2
View solution