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Ch. 4 - Applications of Derivatives
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
All textbooksHass 15th EditionCh. 4 - Applications of DerivativesProblem 23
Chapter 4, Problem 23

The Mean Value Theorem                                                                                                                                                                  
                                                                                                                                                                                        
 a. Show that the equation 𝓍⁴ + 2𝓍² ― 2 = 0 has exactly one solution on [0,1] .
         
[Technology Exercises] b.Find the solution to as many decimal places as you can.  

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Step 1: Define the function f(x) = x^4 + 2x^2 - 2. We need to show that this function has exactly one solution in the interval [0,1].
Step 2: Check the endpoints of the interval. Calculate f(0) and f(1). f(0) = 0^4 + 2*0^2 - 2 = -2 and f(1) = 1^4 + 2*1^2 - 2 = 1.
Step 3: Since f(0) < 0 and f(1) > 0, by the Intermediate Value Theorem, there is at least one root in the interval (0,1).
Step 4: To show that there is exactly one solution, check the derivative f'(x) = 4x^3 + 4x. Analyze the sign of f'(x) on the interval [0,1].
Step 5: Since f'(x) = 4x(x^2 + 1) is always positive for x in [0,1], f(x) is strictly increasing on this interval. Therefore, there is exactly one root in [0,1].

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Mean Value Theorem

The Mean Value Theorem (MVT) states that if a function is continuous on a closed interval [a, b] and differentiable on the open interval (a, b), then there exists at least one point c in (a, b) such that the derivative at that point equals the average rate of change of the function over [a, b]. This theorem is fundamental in understanding the behavior of functions and can be used to prove the existence of solutions to equations.
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Intermediate Value Theorem

The Intermediate Value Theorem (IVT) asserts that for any continuous function on a closed interval [a, b], if the function takes on two values f(a) and f(b), then it must also take on every value between f(a) and f(b) at least once. This theorem is crucial for establishing the existence of roots within an interval, which is essential for showing that the equation x⁴ + 2x² - 2 = 0 has a solution in [0, 1].
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Finding Roots of Polynomials

Finding roots of polynomials involves determining the values of x for which the polynomial equals zero. Techniques such as factoring, synthetic division, and numerical methods (like the Newton-Raphson method) can be employed. In the context of the given equation, analyzing the polynomial's behavior on the interval [0, 1] can help confirm the existence and uniqueness of a root, as well as facilitate the approximation of its value.
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Related Practice
Textbook Question

As a result of a heavy rain, the volume of water in a reservoir increased by 1400 acre-ft in 24 hours. Show that at some instant during that period the reservoir’s volume was increasing at a rate in excess of 225,000 gal/min. (An acre-foot is 43,560 ft³, the volume that would cover 1 acre to the depth of 1 ft. A cubic foot holds 7.48 gal.) 

Textbook Question

Applications


Suppose that f(x) = d/dx (1 − √x) and g(x) = d/dx (x + 2).


Find:


∫[f(x) + g(x)] dx

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Textbook Question

In Exercises 9–66, graph the function using appropriate methods from the graphing procedures presented just before Example 9, identifying the coordinates of any local extreme points and inflection points. Then find coordinates of absolute extreme points, if any.

y = (x² - 49) / (x² + 5x - 14)

Textbook Question

In Exercises 9–66, graph the function using appropriate methods from the graphing procedures presented just before Example 9, identifying the coordinates of any local extreme points and inflection points. Then find coordinates of absolute extreme points, if any.

y=1-(x+1)^3

Textbook Question

Finding Functions from Derivatives


Suppose that f(−1) = 3 and that f'(x) = 0 for all x. Must f(x) = 3 for all x? Give reasons for your answer.

Textbook Question

106. Motion Along a Line The graphs in Exercises 105 and 106 show the position s=f(t) of an object moving up and down on a coordinate line. At approximately what times is the (d) When is the acceleration positive? Negative?