Table of contents
- 0. Functions7h 52m
- Introduction to Functions16m
- Piecewise Functions10m
- Properties of Functions9m
- Common Functions1h 8m
- Transformations5m
- Combining Functions27m
- Exponent rules32m
- Exponential Functions28m
- Logarithmic Functions24m
- Properties of Logarithms34m
- Exponential & Logarithmic Equations35m
- Introduction to Trigonometric Functions38m
- Graphs of Trigonometric Functions44m
- Trigonometric Identities47m
- Inverse Trigonometric Functions48m
- 1. Limits and Continuity2h 2m
- 2. Intro to Derivatives1h 33m
- 3. Techniques of Differentiation3h 18m
- 4. Applications of Derivatives2h 38m
- 5. Graphical Applications of Derivatives6h 2m
- 6. Derivatives of Inverse, Exponential, & Logarithmic Functions2h 37m
- 7. Antiderivatives & Indefinite Integrals1h 26m
- 8. Definite Integrals3h 25m
5. Graphical Applications of Derivatives
Concavity
Problem 22
Textbook Question
Does ƒ(x) = (x⁶/2) + (5x⁴/4) - 15x² have any inflection points? If so, identify them.

1
To find inflection points, we need to determine where the second derivative of the function changes sign. Start by finding the first derivative of the function ƒ(x) = (x⁶/2) + (5x⁴/4) - 15x².
Differentiate ƒ(x) to get the first derivative ƒ'(x). Use the power rule: ƒ'(x) = d/dx[(x⁶/2) + (5x⁴/4) - 15x²].
Calculate the first derivative: ƒ'(x) = (6x⁵/2) + (5 * 4x³/4) - 30x = 3x⁵ + 5x³ - 30x.
Now, find the second derivative ƒ''(x) by differentiating ƒ'(x): ƒ''(x) = d/dx[3x⁵ + 5x³ - 30x].
Calculate the second derivative: ƒ''(x) = 15x⁴ + 15x² - 30. Set ƒ''(x) = 0 and solve for x to find potential inflection points. Check the sign change around these points to confirm inflection points.

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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 are points on a curve where the concavity changes, meaning the curve transitions from being concave up to concave down or vice versa. To find inflection points, one must analyze the second derivative of the function. If the second derivative changes sign at a certain point, that point is classified as an inflection point.
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Second Derivative Test
The second derivative test involves taking the second derivative of a function to determine its concavity. If the second derivative is positive, the function is concave up; if negative, it is concave down. This test is crucial for identifying inflection points, as a change in the sign of the second derivative indicates a potential inflection point.
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The Second Derivative Test: Finding Local Extrema
Critical Points
Critical points are values of x where the first derivative of a function is either zero or undefined. These points are important because they can indicate local maxima, minima, or inflection points. To find inflection points, one must first identify critical points and then analyze the second derivative at these points to check for changes in concavity.
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