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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 4.1.48
Chapter 4, Problem 4.1.48

Finding Critical Points


In Exercises 41–50, determine all critical points and all domain endpoints for each function.


g(x) = √(2x − x²)

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1
First, identify the domain of the function g(x) = √(2x − x²). The expression inside the square root, 2x − x², must be greater than or equal to zero. Solve the inequality 2x − x² ≥ 0 to find the domain.
Rewrite the inequality 2x − x² ≥ 0 as x(2 − x) ≥ 0. This is a quadratic inequality, and we can solve it by finding the roots of the equation x(2 − x) = 0, which are x = 0 and x = 2.
Determine the intervals where the inequality holds by testing values in the intervals (−∞, 0), (0, 2), and (2, ∞). The inequality x(2 − x) ≥ 0 is satisfied in the interval [0, 2]. Thus, the domain of g(x) is [0, 2].
Find the critical points by taking the derivative of g(x). First, express g(x) in a form suitable for differentiation: g(x) = (2x − x²)^(1/2). Use the chain rule to differentiate: g'(x) = (1/2)(2x − x²)^(-1/2) * (2 - 2x).
Set the derivative g'(x) equal to zero to find critical points: (1/2)(2x − x²)^(-1/2) * (2 - 2x) = 0. Solve 2 - 2x = 0 to find x = 1. Check if x = 1 is within the domain [0, 2]. Since it is, x = 1 is a critical point. Also, consider the endpoints of the domain, x = 0 and x = 2, as potential critical points.

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Key Concepts

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

Critical Points

Critical points of a function occur where its derivative is zero or undefined. These points are important because they can indicate local maxima, minima, or points of inflection. To find them, take the derivative of the function and solve for the values of x where the derivative equals zero or does not exist.
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Critical Points

Domain of a Function

The domain of a function is the set of all possible input values (x-values) for which the function is defined. For the function g(x) = √(2x − x²), the expression inside the square root must be non-negative, as square roots of negative numbers are not real. Solving 2x − x² ≥ 0 will give the domain of the function.
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Finding the Domain and Range of a Graph

Derivative of a Function

The derivative of a function represents the rate of change of the function with respect to its variable. For g(x) = √(2x − x²), use the chain rule to differentiate. The derivative helps identify critical points and analyze the behavior of the function, such as increasing or decreasing intervals.
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Derivatives of Other Trig Functions
Related Practice
Textbook Question

Sketch the graphs of the rational functions in Exercises 53–60.


𝓍⁴ ― 1

y = ------------------

𝓍²

Textbook Question

Initial Value Problems


Solve the initial value problems in Exercises 71–90.


ds/dt = 1 + cos t, s(0) = 4

Textbook Question

Theory and Examples


In Exercises 53 and 54, show that the function has neither an absolute minimum nor an absolute maximum on its natural domain.


y = 3x + tan x

Textbook Question

41. Among all triangles in the first quadrant formed by the x-axis, the y-axis, and tangent lines to the graph of y=3x-x^2, what is the smallest possible area?

Textbook Question

Finding Indefinite Integrals


In Exercises 17–56, find the most general antiderivative or indefinite integral. You may need to try a solution and then adjust your guess. Check your answers by differentiation.


∫(8y − 2 / y¹ᐟ⁴) dy

Textbook Question

Finding Indefinite Integrals


In Exercises 17–56, find the most general antiderivative or indefinite integral. You may need to try a solution and then adjust your guess. Check your answers by differentiation.


∫(1 + tan²θ)dθ (Hint:1 + tan²θ = sec²θ)