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Chapter 6: Problem 11

Find the absolute maximum and minimum values of \(f\) on the given closed interval, and state where those values occur. $$f(x)=x-\tan x ;[-\pi / 4, \pi / 4]$$

Short Answer

Expert verified
The absolute maximum is \(f(-\pi/4) = 1 - \pi/4\) at \(-\pi/4\) and the absolute minimum is \(f(\pi/4) = \pi/4 - 1\) at \(\pi/4\).

Step by step solution

01

Finding Critical Points

Find the derivative of the function to locate critical points. The function is given as \( f(x) = x - \tan x \). The derivative, \( f'(x) = 1 - \sec^2 x \), is used to find critical points. Set the derivative equal to zero: \( 1 - \sec^2 x = 0 \), which simplifies to \( \sec^2 x = 1 \). Recall that \( \sec^2 x = 1/\cos^2 x \), meaning \( \cos^2 x = 1 \). This occurs at \( x = 0 \) within the interval \([-\pi/4, \pi/4] \).
02

Evaluate Function at Critical Points

Evaluate the original function at the critical point. Since we found \( x = 0 \) as a critical point, calculate \( f(0) = 0 - \tan(0) = 0 \).
03

Evaluate Function at Endpoints

Evaluate the function \( f(x) = x - \tan x \) at the endpoints of the interval. For \( x = -\pi/4 \), compute \( f(-\pi/4) = -\pi/4 - \tan(-\pi/4) = -\pi/4 + 1 = 1 - \pi/4 \). For \( x = \pi/4 \), calculate \( f(\pi/4) = \pi/4 - \tan(\pi/4) = \pi/4 - 1 = \pi/4 - 1 \).
04

Determine Maximum and Minimum Values

Compare the function values at the critical point and endpoints to identify the absolute maximum and minimum. The values are \( f(0) = 0 \), \( f(-\pi/4) = 1 - \pi/4 \), and \( f(\pi/4) = \pi/4 - 1 \). Convert \( 1 - \pi/4 \) and \( \pi/4 - 1 \) to decimal approximations for clarity: \( f(-\pi/4) \approx 0.215 \) and \( f(\pi/4) \approx -0.215 \). Thus, the absolute maximum value is approximately \( 0.215 \) at \( x = -\pi/4 \), and the absolute minimum value is approximately \( -0.215 \) at \( x = \pi/4 \).

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Key Concepts

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

Critical Points
In calculus, critical points are crucial for finding where a function may have a local maximum or minimum. These points occur where the derivative of the function is zero or undefined. For this exercise, we looked at the function \( f(x) = x - \tan x \). To find critical points, we computed its derivative:
  • \( f'(x) = 1 - \sec^2 x \)
We set \( f'(x) = 0 \) and solved for \( x \). The equation simplifies to \( \sec^2 x = 1 \). Knowing that \( \sec^2 x = 1/\cos^2 x \), we deduce \( \cos^2 x = 1 \). This implies \( \cos x = \pm 1 \), leading us to \( x = 0 \) as a critical point within the interval \([-\pi/4, \pi/4]\).
Derivative
Derivatives are fundamental in calculus, representing the rate of change of a function. They help determine the behavior of functions like increasing, decreasing, and concavity. In this case, the function \( f(x) = x - \tan x \) required finding its derivative:
  • \( f'(x) = 1 - \sec^2 x \)
This derivative gives us insight into the slope of the tangent line at any point on \( f(x) \). Particularly, by setting \( f'(x) = 0 \), we located the critical point. Derivatives also serve to find maximum and minimum values of functions when evaluated at endpoints and critical points.
Absolute Maximum and Minimum
Finding the absolute maximum and minimum involves evaluating the function at critical points and endpoints of a given interval. For this function \( f(x) = x - \tan x \), we applied this method over the interval \([-\pi/4, \pi/4]\):
  • Evaluate at critical point: \( f(0) = 0 - \tan(0) = 0 \).
  • Evaluate at endpoints: \( f(-\pi/4) = 1 - \pi/4 \) and \( f(\pi/4) = \pi/4 - 1 \).
Converting these to decimals, \( f(-\pi/4) \approx 0.215 \) and \( f(\pi/4) \approx -0.215 \). Thus, the absolute maximum value is approximately \( 0.215 \) at \( x = -\pi/4 \), and the absolute minimum value is approximately \( -0.215 \) at \( x = \pi/4 \).
Trigonometric Functions
Trigonometric functions like \( \tan(x) \) are periodic and can alter the behavior of functions significantly over intervals. In \( f(x) = x - \tan x \), \( \tan x \) introduces periodic variation in \( f(x) \). To understand the effects of \( \tan x \), consider its key properties:
  • \( \tan(x) \) has a period of \( \pi \) and is undefined at \( \pm \pi/2 \), making it crucial to restrict domains.
  • At \( x = 0 \), \( \tan(0) = 0 \), simplifying evaluations like \( f(0) \).
  • At \( x = \pm \pi/4 \), \( \tan(\pm \pi/4) = \pm 1 \).
These properties help us compute the function's behavior at different points, aiding in the determination of critical points and absolute extrema.

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Most popular questions from this chapter

A firm determines that \(x\) units of its product can be sold daily at \(p\) dollars per unit, where $$ x=1000-p $$ The cost of producing \(x\) units per day is $$ C(x)=3000+20 x $$ (a) Find the revenue function \(R(x)\) (b) Find the profit function \(P(x)\) (c) Assuming that the production capacity is at most 500 units per day. determine how many units the company must produce and sell each day to maximize the profit. (d) Find the maximum profit. (e) What price per unit must be charged to obtain the maximum profit?

Prove that a particle is speeding up if the velocity and acceleration have the same sign, and slowing down if they have opposite signs. [Hint: Let \(r(t)=|v(t)|=\sqrt{v^{2}(t)},\) and find \(\left.r^{\prime}(t) .\right]\)

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 \(t=5\) $$s(t)=\frac{t}{t^{2}+4}, \quad t \geq 0$$

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. $$f(x)=(x-1)^{2}(x+2)^{2} ;(-\infty,+\infty)$$

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 \(t=5\) $$s(t)=t^{4}-4 t+2, \quad t \geq 0$$
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