/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 30 Find formulas for \(f^{\prime}(x... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 30

Find formulas for \(f^{\prime}(x)\) and \(f^{\prime \prime}(x)\) and state the domains of \(f^{\prime}\) and \(f^{\prime \prime}\). $$ f(x)=|x|^{3} $$

Short Answer

Expert verified
The first derivative is \( f'(x) = \begin{cases} 3x^2 & x \geq 0 \ -3x^2 & x < 0 \end{cases} \) with domain \( (-\infty, \infty) \). The second derivative is \( f''(x) = \begin{cases} 6x & x \geq 0 \ -6x & x < 0 \end{cases} \) with domain \( (-\infty, \infty) \).

Step by step solution

01

- Understand the Given Function

The given function is \[ f(x) = |x|^3 \].We need to find the first and second derivatives and state their domains.
02

- Split the Absolute Value Function

Rewrite the function without the absolute value sign by considering its definition over different intervals: \[ f(x) = \begin{cases} x^3 & x \geq 0 \ (-x)^3 = -x^3 & x < 0 \end{cases} \]
03

- Differentiate on the Interval \(x \geq 0\)

For \(x \geq 0\), we have \[ f(x) = x^3 \].The first derivative is \[ f'(x) = 3x^2 \].
04

- Differentiate on the Interval \(x < 0\)

For \(x < 0\), we have \[ f(x) = -x^3 \].The first derivative is \[ f'(x) = -3x^2 \].
05

- Combine the First Derivatives

Given the different derivatives in each interval, the combined first derivative function is: \[ f'(x) = \begin{cases} 3x^2 & x \geq 0 \ -3x^2 & x < 0 \end{cases} \]
06

- Identify the Domain of the First Derivative

The first derivative \[ f'(x) = \begin{cases} 3x^2 & x \geq 0 \ -3x^2 & x < 0 \end{cases} \] is defined for all real numbers. Therefore, the domain of \(f'(x)\) is \( (-\infty, \infty) \).
07

- Differentiate Again for the Second Derivative

Differentiate the first derivative function: For \(x \geq 0\), \[ f''(x) = \frac{d}{dx}(3x^2) = 6x \],and for \(x < 0\), \[ f''(x) = \frac{d}{dx}(-3x^2) = -6x \].
08

- Combine the Second Derivatives

Combine the second derivative functions: \[ f''(x) = \begin{cases} 6x & x \geq 0 \ -6x & x < 0 \end{cases} \]
09

- Identify the Domain of the Second Derivative

The second derivative \[ f''(x) = \begin{cases} 6x & x \geq 0 \ -6x & x < 0 \end{cases} \] is defined for all real numbers. Therefore, the domain of \( f''(x) \) is \( (-\infty, \infty) \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

first derivative
The first derivative of a function, often denoted as \( f'(x) \), represents the rate at which the function's value changes concerning changes in its input value, \(x\). For our function, \( f(x) = |x|^3 \), we need to first differentiate based on the definition of absolute values over different intervals.

When \( x \geq 0 \):
  • \( f(x) = x^3 \)
  • First derivative: \( f'(x) = 3x^2 \)
When \( x < 0 \):
  • \( f(x) = -x^3 \)
  • First derivative: \( f'(x) = -3x^2 \)
The combined first derivative function is:
\[ f'(x) = \begin{cases} 3x^2 & x \geq 0 \ -3x^2 & x < 0 \end{cases} \]

This combination ensures our first derivative is piecewise-defined based on the intervals of \( x \).
second derivative
The second derivative, denoted as \( f''(x) \), represents the rate at which the first derivative's value changes with respect to \( x \). In essence, it shows the curvature or concavity of the graph of \( f(x) \).

For our piecewise function \( f'(x) \), we differentiate once more:

When \( x \geq 0 \):
  • \( f'(x) = 3x^2 \)
  • Second derivative: \( f''(x) = 6x \)
When \( x < 0 \):
  • \( f'(x) = -3x^2 \)
  • Second derivative: \( f''(x) = -6x \)
The combined second derivative function becomes:
\[ f''(x) = \begin{cases} 6x & x \geq 0 \ -6x & x < 0 \end{cases} \]

This also is piecewise-defined, like the first derivative, and it indicates different behaviors of the function for positive and negative \( x \) values.
absolute value function
The absolute value function describes how to handle real numbers, ensuring they are always non-negative. For any real number \( x \):
\[ |x| = \begin{cases} x & x \geq 0 \ -x & x < 0 \end{cases} \]
In our function, it means:
  • For non-negative values (\( x \geq 0 \)), taking the cube of \( x \) directly.
  • For negative values (\( x < 0 \)), applying negation transforming \( x \) to \( -x \) and then cubing.
Thus, \( f(x) = |x|^3 \) divides into different expressions based on the sign of \( x \), ensuring both sides of the axis are mirrored cubically but in opposite directions.
domain
The domain of a function describes all possible input values (\(x\)) for which the function is defined. For our initial function, \( f(x) = |x|^3 \), any real number \( x \) can be used, so the domain is: \[ (-\infty, \infty) \]

When considering the derivative functions, we find the first and second derivatives:
  • First derivative \( f'(x) \): defined for all real numbers, so domain is \( (-\infty, \infty) \).
  • Second derivative \( f''(x) \): also defined for all real numbers, giving a domain of \( (-\infty, \infty) \).
Hence, any value of \( x \) is accepted in both derivations and original function.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Find the derivative of the given function. $$ g(x)=(2 x-5)^{-1}(4 x+3)^{-2} $$

Differentiate the given function by applying the theorems of this section. $$ f(x)=x^{4}-5+x^{-2}+4 x^{-4} $$

An object is moving along a straight line according to the equation of motion \(s=3 t /\left(t^{2}+9\right)\), with \(t \geq 0\), where \(s\) ft is the directed distance of the object from the starting point at \(t \mathrm{sec}\). (a) What is the instantaneous velocity of the object at \(t_{1}\) sec? (b) What is the instantaneous velocity at \(1 \mathrm{sec}\) ? (c) At what time is the instantaneous velocity zero?

Find the slope of the tangent line to the graph at the point \(\left(x_{1}, y_{1}\right) .\) Make a table of values of \(x, y\), and \(m\) at various points on the graph, and include in the table all points where the graph has a horizontal tangent. Draw a sketch of the graph. $$ y=4 x^{3}-13 x^{2}+4 x-3 $$

Find an equation of each of the lines through the point \((-1,2)\) which is a tangent line to the curve \(y=(x-1) /(x+3)\).
See all solutions

Recommended explanations on Math Textbooks

Discrete Mathematics

Read Explanation

Statistics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Geometry

Read Explanation

Decision Maths

Read Explanation

Mechanics Maths

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.