Problem 31 - The Calculus with Analytic Geometry

Question

Let \(f(x)=|x|\) and let \(F\) be defined by $$ F(x)= \begin{cases}-\frac{1}{2} x^{2} & \text { if } x<0 \\ \frac{1}{2} x^{2} & \text { if } x \geq 0\end{cases} $$ Show that \(F\) is an antiderivative of \(f\) on \((-\infty,+\infty)\).

Step-by-Step Solution

Answer

The derivative of \(F(x)\) is \(f(x) = |x|\) on \((-\infty, +\infty)\).

Step 1: Define Antiderivative

An antiderivative of a function is another function whose derivative is the original function. Therefore, we need to show that the derivative of \(F(x)\) equals \(f(x) = |x|\).

Step 2: Differentiate \(F(x)\) for \(x < 0\)

For \(x < 0\), \(F(x) = -\frac{1}{2} x^2\). Calculate the derivative using the power rule: $$F'(x) = \frac{d}{dx} \left( -\frac{1}{2} x^2 \right) = -x$$

Step 3: Differentiate \(F(x)\) for \(x \geq 0\)

For \(x \geq 0\), \(F(x) = \frac{1}{2} x^2\). Calculate the derivative using the power rule:$$F'(x) = \frac{d}{dx} \left( \frac{1}{2} x^2 \right) = x$$

Step 4: Compare \(F'(x)\) with \(f(x)\)

Combine the results from Steps 2 and 3. For \(x < 0\), \(F'(x) = -x\), which is equal to \(|x|\) since \(|x| = -x\) for negative \(x\). For \(x \geq 0\), \(F'(x) = x\), which is also equal to \(|x|\) since \(|x| = x\) for non-negative \(x\). Therefore, \(F'(x) = f(x)\) on \((-\infty, +\infty)\).

Key Concepts

Antiderivative

An antiderivative of a function is another function whose derivative is the original function. In other words, if you differentiate an antiderivative, you should get back the initial function.
For example, if you have a function \( f(x) = |x| \), an antiderivative is a function \( F(x) \) such that when you take the derivative of \( F(x) \), you end up with \( f(x) \).
In our case, we want to prove that \( F(x) \) as defined in the exercise is an antiderivative of \( f(x) = |x| \). This involves calculating the derivative of \( F(x) \) and showing it equals to \( |x| \) over the entire interval \( ( -\infty, +fty) \).
To achieve this, we need to analyze the function over different ranges of \( x \), specifically where \( x \) is less than 0 and where \( x \) is greater than or equal to 0.

Absolute Value Function

The absolute value function, denoted by \( |x| \), is a function that returns the positive magnitude of a number regardless of its sign. This means:

• \( |x| = x \) if \( x \geq 0 \)
• \( |x| = -x \) if \( x < 0 \)

This definition is crucial when working with piecewise functions like ourselves. Since \( f(x) = |x| \), we have two distinct expressions depending on the sign of \( x \).
For negative \( x \), \( f(x) = |x| = -x \). For non-negative \( x \), \( f(x) = |x| = x \). This behavior must be reflected in the derivatives of the piecewise function \( F(x) \). So, when verifying the antiderivative relationship, we should expect the derivative's result to map back properly to \( |x| \) over the complete domain of \( x \).

Differentiation

Differentiation is the process of computing the derivative of a function. The derivative of a function at a point gives the slope of the tangent line to the function at that point. When we talk about piecewise functions like \( F(x) \), we need to differentiate each piece separately.
Let's revisit our given function \(F(x) = \begin{cases} -\frac{1}{2} x^{2} & \text{if} \( x<0 \) \ \frac{1}{2} x^{2} & \text{if} \( x \geq 0 \) \end{cases} \):

• For \( x < 0 \), \( F(x) = -\frac{1}{2} x^2 \). Using the power rule, we find the derivative: \( F'(x) = \frac{d}{dx} \big( -\frac{1}{2} x^2 \big) = -x \)
• For \( x \geq 0 \), \( F(x) = \frac{1}{2} x^2 \). Similarly, we find the derivative: \( F'(x) = \frac{d}{dx} \big( \frac{1}{2} x^2 \big) = x \)

Combine these results:

• For \( x < 0 \), \( F'(x) = -x \) which matches \( |x| \) since \( |x| = -x \) for negative \( x \).
• For \( x \geq 0 \), \( F'(x) = x \) which matches \( |x| \) since \( |x| = x \) for non-negative \( x \).

Therefore, the derivative \( F'(x) \) is indeed equal to our original function \( f(x) = |x| \) over the entire domain \( ( -\infty, +\infty) \).