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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 123
Chapter 4, Problem 123

In Exercises 121–124, find the inflection points (if any) on the graph of the function and the coordinates of the points on the graph where the function has a local maximum or local minimum value. Then graph the function in a region large enough to show all these points simultaneously. Add to your picture the graphs of the function’s first and second derivatives. How are the values at which these graphs intersect the x-axis related to the graph of the function? In what other ways are the graphs of the derivatives related to the graph of the function?
123. y=(4/5)x^5+16x^2-25

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To find the inflection points, we need to determine where the second derivative of the function changes sign. Start by finding the first derivative of the function \( y = \frac{4}{5}x^5 + 16x^2 - 25 \). Use the power rule to differentiate: \( y' = \frac{d}{dx}(\frac{4}{5}x^5) + \frac{d}{dx}(16x^2) - \frac{d}{dx}(25) \).
Calculate the first derivative: \( y' = \frac{4}{5} \cdot 5x^4 + 32x = 4x^4 + 32x \).
Next, find the second derivative by differentiating the first derivative: \( y'' = \frac{d}{dx}(4x^4 + 32x) \). Apply the power rule again: \( y'' = 16x^3 + 32 \).
To find the inflection points, solve \( y'' = 0 \) for \( x \). Set \( 16x^3 + 32 = 0 \) and solve for \( x \).
To find local maxima and minima, use the first derivative test. Set \( y' = 0 \) and solve for \( x \). This will give you the critical points. Then, use the second derivative test to determine the nature of these critical points: if \( y'' > 0 \), it's a local minimum; if \( y'' < 0 \), it's a local maximum.

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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 is determined by the second derivative. To find inflection points, set the second derivative equal to zero and solve for x. Verify a change in concavity by checking the sign of the second derivative around these points.
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Critical Points

Local Maximum and Minimum

Local maxima and minima are points where a function reaches a highest or lowest value in a neighborhood. These can be found using the first derivative test, where the derivative changes sign, or the second derivative test, where the second derivative is positive (minimum) or negative (maximum) at critical points.
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The Second Derivative Test: Finding Local Extrema

Graphing Derivatives

The first derivative of a function represents its slope, and where it intersects the x-axis indicates critical points. The second derivative indicates concavity, and its x-axis intersections suggest potential inflection points. Graphing these derivatives alongside the original function helps visualize these relationships and understand the function's behavior.
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Graphing The Derivative
Related Practice
Textbook Question

Sketch the graph of a twice-differentiable function y=f(x) that passes through the points (-2,2), (-1,1), (0,0),(1,1), and (2,2) and whose first two derivatives have the following sign patterns.

Textbook Question

Graph f(x) = x cos x and its second derivative together for 0 ≤ x ≤ 2pi. Comment on the behavior of the graph of f in relation to the signs and values of f".

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Textbook Question

Graph f(x) = 2x^4 -4x^2 + 1 and its first two derivatives together. Comment on the behavior of f in relation to the signs and values of f' and f".

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 (b) velocity equal to zero?

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Textbook Question

103. A function f(x) has domain (-2, 2). The graph below is a plot of the derivative of f, not a plot of f itself. In other words, this is a graph of y = f'(x). Either use this graph to determine on which intervals the graph of f is concave up and on which intervals the graph of f is concave down, or explain why this information cannot be determined from the graph.

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