Problem 29 - The Calculus with Analytic Geometry

Question

Find formulas for \(f^{\prime}(x)\) and \(f^{\prime \prime}(x)\) and state the domains of \(f^{\prime}\) and \(f^{\prime \prime}\). $$ f(x)= \begin{cases}-x^{2} & \text { if } x<0 \\ x^{2} & \text { if } x \geq 0\end{cases} $$

Step-by-Step Solution

Answer

The derivatives are: \[f^{\text{'}}(x) = \begin{cases}-2x & \text{if } x < 0 \ 2x & \text{if } x \geq 0 \end{cases}\]\[f^{\text{''}}(x) = \begin{cases}-2 & \text{if } x < 0 \ 2 & \text{if } x \geq 0 \end{cases}\]Domains: \(\mathbb{R}\)

Step 1: Find the first derivative for each case

For the given piecewise function, derive each part separately:For \(x < 0\):\[ f(x) = -x^2 \]\[ f^{\text{'}(x)} = \frac{d}{dx}(-x^2) = -2x \]For \(x \geq 0\):\[ f(x) = x^2 \]\[ f^{\text{'}(x)} = \frac{d}{dx}(x^2) = 2x \]

Step 2: Combine the first derivatives

Combine the first derivatives into a piecewise function:\[ f^{\text{'}(x)} = \begin{cases}-2x & \text{if } x < 0 \ 2x & \text{if } x \geq 0 \end{cases} \]

Step 3: Find the second derivative for each case

For each part of the piecewise function, find the second derivative:For \(x < 0\):\[ f^{\text{'}(x)} = -2x \]\[ f^{\text{''}}(x) = \frac{d}{dx}(-2x) = -2 \]For \(x \geq 0\):\[ f^{\text{'}(x)} = 2x \]\[ f^{\text{''}}(x) = \frac{d}{dx}(2x) = 2 \]

Step 4: Combine the second derivatives

Combine the second derivatives into a piecewise function:\[ f^{\text{''}}(x) = \begin{cases}-2 & \text{if } x < 0 \ 2 & \text{if } x \geq 0 \end{cases} \]

Step 5: State the domains of \(f^{\text{'}}(x)\) and \(f^{\text{''}}(x)\)

For both \(f^{\text{'}}(x)\) and \(f^{\text{''}}(x)\), the domains are all real numbers \(\mathbb{R}\). Since both pieces of the functions \(f^{\text{'}}(x)\) and \(f^{\text{''}}(x)\) are defined for all real numbers, we can conclude that their domains are the same.

Key Concepts

first derivative

Understanding the first derivative of a piecewise function is crucial as it represents the rate of change of the function. For our given piecewise function:
\( f(x)\) = \begin{cases} -x^2 & \text{if } x < 0 \ x^2 & \text{if } x \geq 0 \end{cases},
we first need to derive each part separately.
For \( x < 0 \):
f(x) = -x^2.
The first derivative here is: \( f^{'}(x) = -2x \).
For \( x \geq 0 \):
f(x) = x^2.
The first derivative here is: \( f^{'}(x) = 2x \).
Combining these results, we get the first derivative of the piecewise function:
\( f^{'}(x)\) = \begin{cases} -2x & \text{if } x < 0 \ 2x & \text{if } x \geq 0 \end{cases}.
The first derivative tells us how the function changes. For \( x < 0 \), the function is decreasing, and for \( x \geq 0 \), it is increasing. Understanding the first derivative helps us see the function's behavior and trends in different intervals.

second derivative

The second derivative provides us with information about the curvature or concavity of a function. It highlights whether the function is concave up or concave down. When dealing with our piecewise function:
\( f(x)\) = \begin{cases} -x^2 & \text{if } x < 0 \ x^2 & \text{if } x \geq 0 \end{cases},
we first need to reapply the differentiation process to the first derivative we found:
For \( x < 0 \):
\( f^{'(x)} = -2x \).
The second derivative here is: \( f^{''(x)} = -2 \).
For \( x \geq 0 \):
\( f^{'(x)} = 2x \).
The second derivative here is: \( f^{''(x)} = 2 \).
Combining these, we have the second derivative as:
\( f^{''(x)}\) = \begin{cases} -2 & \text{if } x < 0 \ 2 & \text{if } x \geq 0 \end{cases}.
This information tells us that the function is concave down for \( x < 0 \) and concave up for \( x \geq 0 \). The change in concavity usually suggests an inflection point, although in this case, the drastic switch between segments doesn't provide a classic smooth inflection at the origin. Understanding the second derivative adds depth to the behavior analysis of the function.

domain of a function

Determining the domain of the first and second derivatives is essential as it tells us where these derivatives are valid. For the given piecewise function:
\( f(x)\) = \begin{cases} -x^2 & \text{if } x < 0 \ x^2 & \text{if } x \geq 0 \end{cases},
we identified the first derivative as:
\( f^{'}(x)\) = \begin{cases} -2x & \text{if } x < 0 \ 2x & \text{if } x \geq 0 \end{cases}.
And the second derivative as:
\( f^{''(x)}\) = \begin{cases} -2 & \text{if } x < 0 \ 2 & \text{if } x \geq 0 \end{cases}.
Since both portions of these derivatives are defined for all real numbers, our derivatives are also valid for all real numbers. Therefore, the domain of both \( f^{'}(x) \) and \( f^{''(x)} \) is all real numbers, represented as \( \mathbb{R} \).
Understanding the domain helps us know the extent to which we can apply our derivative functions without issues of undefined values or discontinuities. It's crucial for recognizing the overall scope and authenticity of our analysis.