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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 4.4.8
Chapter 4, Problem 4.4.8

Identify the inflection points and local maxima and minima of the functions graphed in Exercises 1–8. Identify the open intervals on which the functions are differentiable and the graphs are concave up and concave down.
8. y = 2cosx - √2x, -π≤x≤3π/2

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1
To find the inflection points, local maxima, and minima, we first need to find the first and second derivatives of the function y = 2cos(x) - √2x.
The first derivative, y', is found by differentiating each term: y' = -2sin(x) - √2.
The second derivative, y'', is found by differentiating y': y'' = -2cos(x).
To find critical points (potential maxima and minima), set the first derivative y' = 0 and solve for x: -2sin(x) - √2 = 0.
To find inflection points, set the second derivative y'' = 0 and solve for x: -2cos(x) = 0. Analyze the sign changes of y'' to determine intervals of concavity.

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Key Concepts

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

Inflection Points

Inflection points occur where the concavity of a function changes, which can be identified by finding where the second derivative equals zero or is undefined. At these points, the graph transitions from concave up to concave down or vice versa. For the function y = 2cosx - √2x, analyzing the second derivative will help locate these points within the given interval.
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Critical Points

Local Maxima and Minima

Local maxima and minima are points where a function reaches a peak or a trough within a certain interval. These can be found by setting the first derivative to zero and analyzing the sign changes. For y = 2cosx - √2x, identifying these points involves examining the critical points where the derivative changes sign, indicating a shift from increasing to decreasing or vice versa.
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The First Derivative Test: Finding Local Extrema

Concavity and Differentiability

Concavity describes the direction a graph curves, either upwards or downwards, determined by the sign of the second derivative. Differentiability refers to the function having a derivative at all points in an interval. For y = 2cosx - √2x, determining where the graph is concave up or down involves analyzing the second derivative, while differentiability requires checking the continuity and smoothness of the function across the interval.
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Determining Concavity Given a Function
Related Practice
Textbook Question

10. Catching rainwater A 1125 ft^3 open-top rectangular tank with a square base x ft on a side and y ft deep is to be built with its top flush with the ground to catch runoff water. The costs associated with the tank involve not only the material from which the tank is made but also an excavation charge proportional to the product xy.

a. If the total cost is c=5(x^2+4xy) + 10xy, what values of x and y will minimize it?

b. Give a possible scenario for the cost function in part (a).

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Textbook Question

Finding Extrema from Graphs


In Exercises 7–10, find the absolute extreme values and where they occur.


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Textbook Question

Each of Exercises 67–88 gives the first derivative of a continuous function y=f(x). Find y'' and then use Steps 2–4 of the graphing procedure described in this section to sketch the general shape of the graph of f.

82. y' = sin t, for 0 ≤ t ≤ 2π

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Textbook Question

Initial Value Problems


Solve the initial value problems in Exercises 71–90.


dr/dθ = −π sin (πθ), r(0) = 0

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Textbook Question

Absolute Extrema on Finite Closed Intervals


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


g(θ) = θ³ᐟ⁵, −32 ≤ θ ≤ 1

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Textbook Question

Motion with constant acceleration The standard equation for the position s of a body moving with a constant acceleration a along a coordinate line is s = (a/2)t² + v₀t + s₀, where v₀ and s₀ are the body’s velocity and position at time t = 0. Derive this equation by solving the initial value problem

Differential equation: d²s/dt² = a

Initial conditions: ds/dt = v₀ and s = s₀ when t=0.

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