Skip to main content
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 4.2.15aiv
Chapter 4, Problem 4.2.15aiv

Roots (Zeros)


a. Plot the zeros of each polynomial on a line together with the zeros of its first derivative.


iv. y = x³ − 33x² + 216x = x(x - 9)(x − 24)

Verified step by step guidance
1
Identify the zeros of the polynomial y = x³ − 33x² + 216x. The polynomial is factored as y = x(x - 9)(x - 24), so the zeros are x = 0, x = 9, and x = 24.
Find the first derivative of the polynomial y = x³ − 33x² + 216x. Use the power rule for differentiation: if y = ax^n, then dy/dx = n*ax^(n-1).
Apply the power rule to each term: the derivative of x³ is 3x², the derivative of -33x² is -66x, and the derivative of 216x is 216. Therefore, the first derivative is y' = 3x² - 66x + 216.
Find the zeros of the first derivative y' = 3x² - 66x + 216. Set the derivative equal to zero: 3x² - 66x + 216 = 0. This is a quadratic equation.
Solve the quadratic equation 3x² - 66x + 216 = 0 using the quadratic formula x = (-b ± √(b² - 4ac)) / 2a, where a = 3, b = -66, and c = 216. Calculate the discriminant and find the roots.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.
Video duration:
2m
Was this helpful?

Key Concepts

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

Roots (Zeros) of a Polynomial

The roots or zeros of a polynomial are the values of x for which the polynomial equals zero. For the polynomial y = x³ − 33x² + 216x, the roots are found by setting y to zero and solving the equation x(x - 9)(x - 24) = 0, giving the roots x = 0, x = 9, and x = 24. These points are where the graph of the polynomial intersects the x-axis.
Recommended video:
6:04
Introduction to Polynomial Functions

First Derivative and Critical Points

The first derivative of a function, denoted as y', represents the rate of change or slope of the function. For y = x³ − 33x² + 216x, the first derivative is y' = 3x² - 66x + 216. The zeros of the first derivative, found by solving 3x² - 66x + 216 = 0, indicate critical points where the function's slope is zero, potentially corresponding to local maxima, minima, or points of inflection.
Recommended video:
04:50
Critical Points

Graphical Representation of Zeros

Plotting the zeros of a polynomial and its derivative on a number line helps visualize the relationship between the function and its rate of change. The zeros of the polynomial indicate where the function crosses the x-axis, while the zeros of the derivative show where the slope is zero. This graphical approach aids in understanding the function's behavior, such as identifying intervals of increase or decrease and locating turning points.
Recommended video:
Guided course
6:37
Zero and Negative Rules
Related Practice
Textbook Question

Roots (Zeros)


a. Plot the zeros of each polynomial on a line together with the zeros of its first derivative.


iii. y = x³ − 3x² + 4 = (x + 1)(x − 2)²

1
views
Textbook Question

Identifying Extrema


In Exercises 41–52:


a. Identify the function’s local extreme values in the given domain, and say where they occur.


g(x) = (x − 2) / (x²−1), 0 ≤ x < 1

Textbook Question

Roots (Zeros)


a. Plot the zeros of each polynomial on a line together with the zeros of its first derivative.


i. y = x² − 4

Textbook Question

Theory and Examples


In Exercises 51 and 52, give reasons for your answers.


Let f(x) = (x − 2)²ᐟ³.


b. Show that the only local extreme value of f occurs at x = 2.

Textbook Question

Finding Antiderivatives

In Exercises 1–16, find an antiderivative for each function. Do as many as you can mentally. Check your answers by differentiation.

-(1/2)x⁻³ᐟ²

1
views
Textbook Question

Theory and Examples


Cubic functions Consider the cubic function f(x) = ax³ + bx² + cx + d.


b. How many local extreme values can f have?