Problem 1
Question
In Problems \(1-12,\) a function \(f\) and its domain are given. Determine the critical points, evaluate \(f\) at these points, and find the (global) maximum and minimum values. \(f(x)=x^{2}-2 x ;[0,4]\)
Step-by-Step Solution
Verified Answer
The global maximum is 8 at \(x = 4\) and the global minimum is -1 at \(x = 1\).
1Step 1: Identify the Derivative
To find the critical points, we need to find where the derivative of the function is zero or undefined. First, find the derivative of the function. Given the function, \( f(x) = x^2 - 2x \). Use the power rule to differentiate.The derivative is: \( f'(x) = 2x - 2 \).
2Step 2: Solve for Critical Points
Set the derivative equal to zero to find the critical points.So, \( 2x - 2 = 0 \).Solve for \(x\):\[ 2x = 2 \ x = 1 \].This means there is one critical point at \( x = 1 \).
3Step 3: Evaluate at Critical Points and Endpoints
Evaluate the function \( f(x) \) at the critical point and at the endpoints of the domain [0, 4].Calculate \( f(1) \):\( f(1) = 1^2 - 2 imes 1 = 1 - 2 = -1 \).Next, calculate at the endpoints:\( f(0) = 0^2 - 2 imes 0 = 0 \).\( f(4) = 4^2 - 2 imes 4 = 16 - 8 = 8 \).
4Step 4: Determine Maximum and Minimum
Compare the function values from the critical point and at the endpoints to determine the global maximum and minimum on the interval [0, 4].The values are:\( f(0) = 0 \)\( f(1) = -1 \)\( f(4) = 8 \).The maximum value is 8, occurring at \(x = 4\), and the minimum value is -1, occurring at \(x = 1\).
Key Concepts
DerivativeInterval EvaluationGlobal Maximum and Minimum
Derivative
The concept of a derivative is central to calculus. It helps us understand the rate of change of a function at any given point. In simpler terms, the derivative tells us the slope of a function's graph at a specific point.
To find the derivative of a function like \( f(x) = x^2 - 2x \), we use differentiation rules, such as the power rule. In this case, the derivative \( f'(x) \) is calculated as \( f'(x) = 2x - 2 \).
This derivative allows us to find critical points by setting \( f'(x) = 0 \), providing values of \( x \) where the function could be maximized or minimized.
To find the derivative of a function like \( f(x) = x^2 - 2x \), we use differentiation rules, such as the power rule. In this case, the derivative \( f'(x) \) is calculated as \( f'(x) = 2x - 2 \).
This derivative allows us to find critical points by setting \( f'(x) = 0 \), providing values of \( x \) where the function could be maximized or minimized.
Interval Evaluation
Interval evaluation involves considering the values of a function over a specified range, or interval, of \( x \). In this problem, the interval given is \([0, 4]\).
The first step is to evaluate the function at the critical points found from the derivative, which gives potential places where a maximum or minimum might occur.
Next, evaluate the function at the endpoints of the interval as well. This is because the absolute extremes (global maximum or minimum values) of a continuous function on a closed interval will occur either at critical points or at the interval's endpoints.
In our example, we evaluate \( f(x) \) at the critical point \( x = 1 \), and at the endpoints \( x = 0 \) and \( x = 4 \).
The first step is to evaluate the function at the critical points found from the derivative, which gives potential places where a maximum or minimum might occur.
Next, evaluate the function at the endpoints of the interval as well. This is because the absolute extremes (global maximum or minimum values) of a continuous function on a closed interval will occur either at critical points or at the interval's endpoints.
In our example, we evaluate \( f(x) \) at the critical point \( x = 1 \), and at the endpoints \( x = 0 \) and \( x = 4 \).
Global Maximum and Minimum
Once you have evaluated the function at both the critical points and the interval endpoints, you can determine the global maximum and minimum values.
The global maximum is the highest function value within the interval, and the global minimum is the lowest value.
In this example, after evaluating \( f(x) \) at \( x = 0, 1, \) and \( 4 \), we find the values are \( f(0) = 0 \), \( f(1) = -1 \), and \( f(4) = 8 \).
Here, the global maximum is \( 8 \) at \( x = 4 \), and the global minimum is \(-1\) at \( x = 1 \).
By comparing these values, we ensure that we've identified the entire range of the function's behavior over the given interval.
The global maximum is the highest function value within the interval, and the global minimum is the lowest value.
In this example, after evaluating \( f(x) \) at \( x = 0, 1, \) and \( 4 \), we find the values are \( f(0) = 0 \), \( f(1) = -1 \), and \( f(4) = 8 \).
Here, the global maximum is \( 8 \) at \( x = 4 \), and the global minimum is \(-1\) at \( x = 1 \).
By comparing these values, we ensure that we've identified the entire range of the function's behavior over the given interval.
Other exercises in this chapter
Problem 1
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