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Chapter 4: Problem 19

In Exercises \(15-30\) , 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. $$ F(x)=-\frac{1}{x^{2}}, \quad 0.5 \leq x \leq 2 $$

Short Answer

Expert verified
Absolute minimum is -4 at x = 0.5; absolute maximum is -0.25 at x = 2.

Step by step solution

01

Verify the Given Function and Interval

The function provided is \( F(x) = -\frac{1}{x^2} \) and the interval is \( 0.5 \leq x \leq 2 \). Our task is to find the absolute maximum and minimum values of \( F(x) \) within this interval.
02

Find the Critical Points

First, we find the derivative of the function: \[ F'(x) = \frac{d}{dx}\left(-\frac{1}{x^2}\right) = \frac{2}{x^3} \] Now set the derivative equal to zero to find critical points: \[ \frac{2}{x^3} = 0 \] Since this equation has no solutions, there are no critical points inside the interval \( (0.5, 2) \).
03

Evaluate the Function at the Endpoints

Since there are no critical points within the interval, we evaluate \( F(x) \) at the endpoints of the interval: \[ F(0.5) = -\frac{1}{(0.5)^2} = -4 \] \[ F(2) = -\frac{1}{2^2} = -0.25 \]
04

Determine Absolute Extrema

The value of \( F(x) \) at \( x = 0.5 \) is \(-4\), and at \( x = 2 \) it is \(-0.25\). Thus, within the interval \(0.5 \leq x \leq 2\), the absolute minimum is \(-4\) and it occurs at \( x = 0.5 \), while the absolute maximum is \(-0.25\) and it occurs at \( x = 2 \).
05

Graph the Function and Identify Extrema Points

Plot the function \( F(x) = -\frac{1}{x^2} \) on the interval \( 0.5 \leq x \leq 2 \). Mark the points \((0.5, -4)\) and \((2, -0.25)\) on the graph. - The absolute minimum \((-4)\) is at \(x = 0.5\).- The absolute maximum \((-0.25)\) is at \(x = 2\).

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Key Concepts

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

Derivative
The derivative is a mathematical tool we use to understand how a function changes as its input changes. It tells us the slope of a function at any given point. In simpler terms, it shows us how fast or slow the function is rising or falling.
To find the derivative of the function, we use the power rule or other differentiation techniques. For the function given, \( F(x) = -\frac{1}{x^2} \), we apply the rule for deriving a power: if \( y = x^n \), then \( \frac{dy}{dx} = nx^{n-1} \).
Thus, the derivative \( F'(x) = \frac{d}{dx}(-x^{-2}) \) becomes \( \frac{2}{x^3} \).
  • Helps identify the behavior of the function, such as increasing or decreasing.
  • Essential in finding critical points where extrema might occur.
Critical Points
Critical points of a function are where its derivative is either zero or undefined. At these points, the function can have a local maximum, minimum, or a saddle point.
To find the critical points, we set the derivative equal to zero: \( \frac{2}{x^3} = 0 \).

Since this equation has no solutions, there are no critical points for the function \( F(x) = -\frac{1}{x^2} \) within the interval \( (0.5, 2) \).
  • Critical points help determine potential extrema inside an interval.
  • Absent critical points mean checking endpoints for extrema.
Endpoints
Endpoints refer to the boundary values of a given interval. If critical points are absent, as in our case, we need to check the endpoints to find absolute extrema.
The endpoints for our interval \( [0.5, 2] \) are \( x = 0.5 \) and \( x = 2 \). We evaluate the function at these endpoints:
  • \( F(0.5) = -\frac{1}{(0.5)^2} = -4 \)
  • \( F(2) = -\frac{1}{2^2} = -0.25 \)
By comparing these values, we identify the minimum and maximum function values on the interval.
Interval
An interval is a range of values, typically denoted as \( [a, b] \) where \( a \) and \( b \) are the endpoints. Our task is often to find the output of a function within this specific range.
In the exercise, the interval is \( 0.5 \leq x \leq 2 \). This simply means we only focus on values of \( x \) that lie within, and including, these boundaries.
  • Determines where we look for critical points and endpoints.
  • Helps limit the domain for considering the absolute extrema.
Understanding intervals is crucial when analyzing where extrema happen since only candidate points inside these boundaries are considered valid.

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