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Chapter 3: Problem 21

Find the number in the interval [0,3] such that the number minus its square is: a. As large as possible. b. As small as possible.

Short Answer

Expert verified
The number for maximum value is \( \frac{1}{2} \) and for minimum is 3.

Step by step solution

01

Define the Function

First, define the function that represents the expression given in the problem. Let \( f(x) = x - x^2 \). This function will help us understand the relationship between \( x \) and \( x^2 \).
02

Find the Derivative

Differentiate the function \( f(x) = x - x^2 \) to find the critical points. The derivative is \( f'(x) = 1 - 2x \).
03

Solve for Critical Points

Set the derivative \( f'(x) = 0 \) to find the critical points. Solve \( 1 - 2x = 0 \), which gives \( x = \frac{1}{2} \).
04

Evaluate the Function at Critical Points and Endpoints

Substitute the critical point \( x = \frac{1}{2} \) and the endpoints \( x = 0 \) and \( x = 3 \) into the function \( f(x) = x - x^2 \).- \( f(0) = 0 - 0^2 = 0 \)- \( f(\frac{1}{2}) = \frac{1}{2} - (\frac{1}{2})^2 = \frac{1}{2} - \frac{1}{4} = \frac{1}{4} \)- \( f(3) = 3 - 3^2 = 3 - 9 = -6 \)
05

Determine Maximum and Minimum Values

Compare the values of \( f(x) \) at the critical point and endpoints:- Maximum value occurs at \( x = \frac{1}{2} \) with \( f(\frac{1}{2}) = \frac{1}{4} \).- Minimum value occurs at \( x = 3 \) with \( f(3) = -6 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding the Derivative
A derivative helps us understand how a function changes at any point. It tells us the slope of the tangent line to the function curve at a given point. This concept is vital when you want to find the peaks or valleys of a function, known as critical points. For the function \( f(x) = x - x^2 \), the derivative is \( f'(x) = 1 - 2x \).
To find critical points, we solve \( f'(x) = 0 \). By doing this for \( 1 - 2x = 0 \), we find that the critical point is at \( x = \frac{1}{2} \). This point potentially represents where the function reaches a maximum or minimum within the given interval.
Function Evaluation
Function evaluation involves substituting specific values into a function to get outputs. In our exercise, we evaluate the function \( f(x) = x - x^2 \) at the critical point and the endpoints of the interval [0,3].
  • At \( x = 0 \), \( f(0) = 0 - 0^2 = 0 \).
  • At the critical point \( x=\frac{1}{2} \), \( f(\frac{1}{2}) = \frac{1}{2} - \left(\frac{1}{2}\right)^2 = \frac{1}{4} \).
  • At \( x = 3 \), \( f(3) = 3 - 3^2 = -6 \).
By evaluating these points, you can determine how the function behaves over the interval and identify the maximum and minimum values.
Exploring Optimization
Optimization is about finding the best possible value of a function under given conditions. This could mean finding the maximum or minimum value. In our exercise, the goal was to find when the expression \( x - x^2 \) is largest and smallest within the interval [0,3].
  • Maximum: Occurs at \( x = \frac{1}{2} \) with the value \( \frac{1}{4} \).
  • Minimum: Occurs at \( x = 3 \) with the value \( -6 \).
By evaluating the function at critical points and endpoints, we determine where the extremes鈥攎aximum and minimum鈥攁re within the interval. This is the essence of optimization.
Interval Analysis and Its Importance
Interval analysis involves looking at how a function behaves over a specific range of inputs, known as an interval. For the exercise at hand, the interval is [0,3]. By analyzing this interval:
  • We ensure that all evaluations stay within the scope of our interest.
  • It helps us verify that the identified maximum and minimum values are valid within the desired range.
  • It allows us to identify any endpoints, which could also potentially hold the maximum or minimum values.
Using interval analysis, we validate our findings as relevant and correct based on the limits set by this interval.

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Most popular questions from this chapter

True or False: If the derivative has the same sign immediately on either side of an \(x\) -value, the function has neither a maximum nor a minimum at that \(x\) -value.

Find the linear approximation for each root by choosing the closest value of \(x\) for which the calculation is easy and then using the corresponding value of \(\Delta x\). $$ \sqrt{48} $$

The value \(V(t)\) of a case of investment-grade wine after \(t\) years is $$ V(t)=2000+80 \sqrt{t}-10 t $$ dollars (for \(0 \leq t \leq 25\) ). For each number of years, find \(d V\) and use it to estimate the value one year later. $$ \text { 16 years } $$

In general, implicit differentiation gives an expression for the derivative that involves both \(x\) and \(y\). Under what conditions will the expression involve only \(x ?\)

Suppose that you have a formula that relates the amount of gas used (denoted by \(x\) ) to the distance driven (denoted by \(y\) ) in your car. State, in everyday language, what \(\frac{d y}{d x}\) and \(\frac{d x}{d y}\) would mean.
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