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Chapter 6: Problem 15

Find the absolute maximum and minimum values of \(f\), if any, on the given interval, and state where those values occur. $$f(x)=x^{2}-3 x-1 ;(-\infty,+\infty)$$

Short Answer

Expert verified
The function has an absolute minimum value of \(-\frac{13}{4}\) at \(x = \frac{3}{2}\), and no absolute maximum on \((-\infty, +\infty)\).

Step by step solution

01

Find the derivative of the function

To find the critical points of the function, we first find the derivative of \( f(x) = x^2 - 3x - 1 \). The derivative is calculated as follows: \[ f'(x) = \frac{d}{dx}(x^2 - 3x - 1) = 2x - 3 \].
02

Set the derivative to zero to find critical points

To find the critical points, we set the derivative \( f'(x) \) to zero and solve for \( x \):\[ 2x - 3 = 0 \]\[ 2x = 3 \]\[ x = \frac{3}{2} \].This is the critical point of the function.
03

Evaluate the function at critical points and endpoints

Since the interval is \((-\infty,+\infty)\), there are no finite endpoints to evaluate. We evaluate the function at the critical point \( x = \frac{3}{2} \):\[ f\left(\frac{3}{2}\right) = \left(\frac{3}{2}\right)^2 - 3\left(\frac{3}{2}\right) - 1 \]\[ = \frac{9}{4} - \frac{9}{2} - 1 \]\[ = \frac{9}{4} - \frac{18}{4} - \frac{4}{4} \]\[ = \frac{9 - 18 - 4}{4} \]\[ = \frac{-13}{4} \].
04

Determine if critical point is a minimum or maximum

To determine whether the critical point is a maximum or minimum, evaluate the second derivative:\[ f''(x) = \frac{d}{dx}(2x - 3) = 2 \].Since \( f''(x) = 2 \) is positive for all \( x \), the function is concave up, meaning the critical point at \( x = \frac{3}{2} \) is a minimum.

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Key Concepts

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

Calculus
Calculus is a branch of mathematics that helps us understand changes and motion. It provides a framework for analyzing the behavior of functions. When we talk about finding extrema (or extreme values) in calculus, we're typically looking for the highest or lowest points on a graph.
This involves a few important steps such as finding derivatives, establishing critical points, and assessing concavity. Through calculus, we can evaluate how a function behaves and predict where it reaches its maximum or minimum values across the given interval. In our exercise, we worked through a function to find where it hits its lowest point on a continuous interval.
Critical Points
Critical points are where the derivative of a function either equals zero or is undefined. These points are crucial because they can signify where the function changes direction, potentially hitting maximums or minimums.
For instance, if the slope of the tangent line (i.e., the derivative) is zero, it suggests a potential peak or valley. In our exercise, we found the critical point by setting the derivative
  • \( f'(x) = 2x - 3 \)
equal to zero, solving for \( x \), and finding \( x = \frac{3}{2} \).Knowing these critical points allows us to focus on specific spots of interest in the graph, particularly when assessing whether they stand as relative maximum or minimum values.
Concavity
Concavity informs us about the curvature of a graph. When we determine concavity, we're essentially asking whether the curve is opening up or down at a particular point. This is critical in establishing whether critical points are relative maxima or minima.
A function is concave up if its second derivative is positive, meaning the curve opens like a smile :) and suggests minima around critical points. Conversely, if the second derivative is negative, the curve opens downward like a frown :( and suggests maxima.
  • In the solution, we used the second derivative \( f''(x) = 2 \)
to verify that it is concave up everywhere, as this value is consistently positive. Thus, the critical point \( x = \frac{3}{2} \) is a minimum.
Derivatives
Derivatives assess how a function changes at any given point. When we derive a function, we're calculating the rate at which our function's value is changing with respect to its variable. This is symbolized as \( f'(x) \).
The first derivative gives us the slope or rate of change of the original function. By setting the first derivative to zero, we locate the critical points which indicate places where the function might achieve extreme values. In our exercise:
  • The derivative of \( f(x) = x^2-3x-1 \)
  • Was found to be \( f'(x) = 2x-3 \)
Finding derivatives is a key part of determining behavior and identifying the points of interest on a function's graph.

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Most popular questions from this chapter

Use a graphing utility to estimate the absolute maximum and minimum values of \(f\), if any, on the stated interval, and then use calculus methods to find the exact values. $$f(x)=2 \sec x-\tan x ;[0, \pi / 4]$$

Two particles. \(A\) and \(B\), are in motion in the \(x y\) -plane. Their coordinates at each instant of time \(t(t \geq 0)\) are given by \(x_{A}=t, y_{A}=2 t, x_{B}=1-t,\) and \(y_{B}=t .\) Find the minimum distance between \(A\) and \(B\)

(a) Use the Mean-Value Theorem to show that if \(f\) is differentiable on an open interval \(I\), and if \(\left|f^{\prime}(x)\right| \geq M\) for all values of \(x\) in \(I\), then $$|f(x)-f(y)| \geq M|x-y|$$ for all values of \(x\) and \(y\) in \(I\). (b) Use the result in part (a) to show that $$|\tan x-\tan y| \geq|x-y|$$ for all values of \(x\) and \(y\) in the interval \((-\pi / 2, \pi / 2)\). (c) Use the result in part (b) to show that $$|\tan x+\tan y| \geq|x+y|$$ for all values of \(x\) and \(y\) in the interval \((-\pi / 2, \pi / 2)\).

(a) Use the chain rule to show that for a particle in rectilinear motion \(a=v(d v / d s)\) (b) Let \(s=\sqrt{3 t+7}, t \geq 0 .\) Find a formula for \(v\) in terms of \(s\) and use the equation in part (a) to find the acceleration when \(s=5\)

The position function of a particle moving along a coordinate line is given, where \(s\) is in feet and \(t\) is in seconds. (a) Find the velocity and acceleration functions. (b) Find the position, velocity, speed, and acceleration at time \(t=1\) (c) At what times is the particle stopped? (d) When is the particle speeding up? Slowing down? (e) Find the total distance traveled by the particle from time \(t=0\) to time \(t=5\) $$s(t)=\frac{t}{t^{2}+4}, \quad t \geq 0$$
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