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Chapter 2: Problem 69

Find \(f^{\prime \prime}(x)\) if \(f(x)=\left|x^{3}\right|\). Does \(f^{\prime \prime}(0)\) exist?

Short Answer

Expert verified
The second derivative of \(f(x) = |x^3|\) is given by \(f^{\prime\prime}(x) = -6x\) for \(x<0\), and \(f^{\prime\prime}(x) = 6x\) for \(x\geq0\). Since both left and right limits of the second derivative agree at \(x=0\), we can conclude that \(f^{\prime\prime}(0)\) exists and is equal to 0.

Step by step solution

01

Rewrite the function as a piecewise function

First, rewrite the function as a piecewise function considering that: \(|x^3| = -x^3\) for \(x < 0\), and \(|x^3| = x^3\) for \(x \geq 0\). So, we have: \[ f(x) = \begin{cases} -x^3 & , \text{if } x<0 \\ x^3 & , \text{if } x\geq0 \end{cases} \]
02

Find the first derivative of f(x)

Now, calculate the first derivative of the piecewise function: \[ f^{\prime}(x) = \frac{d}{dx}(f(x)) = \begin{cases} -3x^2 & , \text{if } x<0 \\ 3x^2 & , \text{if } x\geq0 \end{cases} \]
03

Find the second derivative of f(x)

Next, calculate the second derivative of the piecewise function: \[f^{\prime\prime}(x) = \frac{d^2}{dx^2}(f(x)) = \begin{cases} -6x & , \text{if } x<0 \\ 6x & , \text{if } x\geq0 \end{cases} \]
04

Determine if \(f^{\prime\prime}(0)\) exists

Finally, we'll determine if the second derivative exists at \(x=0\): For \(x<0\), we have \(f^{\prime\prime}(x) = -6x\), and as x approaches 0 from the left, \(f^{\prime\prime}(x)\) approaches 0. For \(x\geq 0\), we have \(f^{\prime\prime}(x) = 6x\), and as x approaches 0 from the right, \(f^{\prime\prime}(x)\) also approaches 0. Since both the left and right limits of the second derivative agree, we can conclude that \(f^{\prime\prime}(0)\) exists, and it is equal to 0.

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Key Concepts

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

Understanding Piecewise Functions
A piecewise function is a function made up of different expressions depending on the value of the input, typically defined over specific intervals. For example, the absolute value function we have, \(|x^3|\), becomes different functions when \(x\) is less than 0 and when it's greater than or equal to 0. This is because:
  • When \(x < 0\), \(|x^3| = -x^3\)
  • When \(x \geq 0\), \(|x^3| = x^3\)
Thus, the function is written as:\[f(x) = \begin{cases}-x^3 & , \text{if } x<0 \x^3 & , \text{if } x\geq0\end{cases}\]This approach allows us to handle different operations on \(f(x)\) based on where \(x\) falls. It is crucial in calculus because each piece can have its own derivative and behavior.
Finding the First Derivative
The first derivative represents the rate at which the function changes. For a piecewise function, it鈥檚 important to find the derivative for each segment separately. Here, for \(f(x)\), we calculate:
  • For \(x < 0\), \,\( f^{\prime}(x) = -3x^2\)
  • For \(x \geq 0\), \,\( f^{\prime}(x) = 3x^2\)
This gives us:\[f^{\prime}(x) = \begin{cases}-3x^2 & , \text{if } x<0 \3x^2 & , \text{if } x\geq0\end{cases}\]Calculating the first derivative for each piece separately allows us to understand how the function \(f(x)\) behaves differently across its domain.
The Concept of Derivatives at a Point
Derivatives at a point, like \(f^{\prime\prime}(0)\), assess the behavior of the derivative precisely at a single value of \(x\). Determining \(f^{\prime\prime}(0)\) involves examining the behavior of the second derivative from both sides of zero:
  • For \(x < 0\), \,\( f^{\prime\prime}(x) = -6x\)
  • For \(x \geq 0\), \,\( f^{\prime\prime}(x) = 6x\)
As \(x\) approaches 0 from either side, both expressions of the second derivative tend to 0:
  • From the left: \,\( -6x \to 0 \)
  • From the right: \,\( 6x \to 0 \)
Because these limits agree, \(f^{\prime\prime}(0)\) exists and equals 0. This agreement ensures the continuity of the second derivative at that point, confirming its existence. Understanding derivatives at a point helps confirm smoothness and other characteristics of functions at specific points.

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