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Ch. 4 - Applications of Derivatives
Hass - Thomas' Calculus 15th Edition
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모든 교과서Hass 15th EditionCh. 4 - Applications of Derivatives문제 4.1.45
4장, 문제 4.1.45

Finding Critical Points


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


y = x² + 2/x

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First, identify the domain of the function y = x² + 2/x. Since the function includes a term 2/x, x cannot be zero. Therefore, the domain is all real numbers except x = 0.
To find the critical points, we need to find the derivative of the function y = x² + 2/x. Use the power rule and the quotient rule to differentiate: y' = d/dx (x²) + d/dx (2/x).
Calculate the derivative: y' = 2x - 2/x². This involves differentiating x² to get 2x and using the power rule on 2/x to get -2/x².
Set the derivative equal to zero to find critical points: 2x - 2/x² = 0. Solve this equation for x to find the values where the slope of the tangent is zero.
Additionally, check the endpoints of the domain. Since the domain is all real numbers except x = 0, consider the behavior of the function as x approaches zero from both the positive and negative sides to understand the behavior at the domain boundaries.

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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.
추천 영상:
04:50
Critical Points

Derivative

The derivative of a function represents the rate of change of the function with respect to its variable. It is a fundamental tool in calculus used to find slopes of tangent lines, velocities, and other rates of change. For the function y = x² + 2/x, the derivative is found using the power rule and the quotient rule.
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가이드 코스
05:44
Derivatives

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 rational functions like y = x² + 2/x, the domain excludes values that make the denominator zero. Identifying the domain is crucial for understanding where critical points and endpoints can occur.
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가이드 코스
5:10
Finding the Domain and Range of a Graph
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Finding Extreme Values

In Exercises 1–10, find the extreme values (absolute and local) of the function over its natural domain, and where they occur.


y = 𝓍³ ― 2𝓍 + 4

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Identifying Extrema


In Exercises 19–40:


a. Find the open intervals on which the function is increasing and those on which it is decreasing.


b. Identify the function’s local extreme values, if any, saying where they occur.


f(x) = x³ / (3x² + 1)

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"Roots (Zeros) Show that the functions in Exercises 19–26 have exactly one zero

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Absolute Extrema on Finite Closed Intervals


In Exercises 37–40, find the function’s absolute maximum and minimum values and say where they occur.


f(x) = x⁴ᐟ³, −1 ≤ x ≤ 8

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Solve the initial value problems in Exercises 71–90.

y⁽⁴⁾ = −sin t + cos t;

y′′′(0) =7, y′′(0) = y′(0) = −1, y(0) = 0

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Checking the Mean Value Theorem


Find the value or values of c that satisfy the equation (f(b) − f(a)) / (b − a) = f′(c) in the conclusion of the Mean Value Theorem for the functions and intervals in Exercises 1–6.


g(x) = {x³, −2 ≤ x ≤ 0

x², 0 < x ≤ 2

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