Finding the Derivative of a Function with Absolute Values

The problem asks for the derivative of the function \(f(x) = (|x| - |x – 1|)^2\) when \(x > 1\).

To solve this, we first need to simplify the function \(f(x)\) for the specific interval where \(x > 1\). The presence of absolute values requires us to consider the sign of the expressions inside the absolute value signs.

Simplifying the Function for \(x > 1\)

The definition of absolute value is:

• \(|a| = a\) if \(a \ge 0\)
• \(|a| = -a\) if \(a < 0\)

Let's analyze the terms inside the absolute values when \(x > 1\):

• For \(|x|\): Since \(x > 1\), \(x\) is positive. Therefore, \(|x| = x\).
• For \(|x - 1|\): Since \(x > 1\), \(x - 1 > 1 - 1 = 0\). So, \(x - 1\) is positive. Therefore, \(|x - 1| = x - 1\).

Now, substitute these simplified expressions back into the function \(f(x)\) for \(x > 1\):

\[f(x) = (|x| - |x – 1|)^2\]

For \(x > 1\):

\[f(x) = (x - (x - 1))^2\]

Let's simplify the expression inside the parenthesis:

\[x - (x - 1) = x - x + 1 = 1\]

So, for \(x > 1\), the function simplifies to:

\[f(x) = (1)^2\]

\[f(x) = 1\]

Thus, for all values of \(x\) strictly greater than 1, the function \(f(x)\) is simply the constant value 1.

Calculating the Derivative \(f’(x)\)

Now we need to find the derivative of \(f(x)\) with respect to \(x\) when \(x > 1\). We found that \(f(x) = 1\) for \(x > 1\).

The derivative of a constant function is always 0.

So, if \(f(x) = 1\) for \(x > 1\), then the derivative \(f’(x)\) is:

\[f’(x) = \frac{d}{dx}(1)\]

\[f’(x) = 0\]

Therefore, when \(x > 1\), the derivative \(f’(x)\) is equal to 0.

Comparing with the Options

Let's compare our result with the given options:

• Option 1: 0
• Option 2: \(2x - 1\)
• Option 3: \(4x - 2\)
• Option 4: \(8x - 4\)

Our calculated derivative, \(f’(x) = 0\), matches Option 1.

Understanding Absolute Value in Calculus

Functions involving absolute values are often piecewise functions. To find the derivative of such functions, it's crucial to define the function explicitly without the absolute value signs over different intervals. The points where the expressions inside the absolute value become zero are critical points that define these intervals. For \(|x|\), the critical point is \(x=0\). For \(|x-1|\), the critical point is \(x=1\). These points divide the number line into intervals: \(x < 0\), \(0 \le x < 1\), and \(x \ge 1\). We were specifically asked to consider the interval \(x > 1\).

Revision Table: Function Definition and Derivative

Note: The derivative at the critical points (like \(x=0\) and \(x=1\)) might not exist or requires checking limits, but the question asks for the derivative when \(x > 1\), which is within an open interval where the function is smooth.

Additional Information: Derivatives of Piecewise Functions

When dealing with piecewise functions, especially those defined using absolute values, the derivative of each piece can be found using standard differentiation rules. However, the derivative at the points where the definition of the function changes (the critical points) needs special attention. A function must be continuous at a point for the derivative to exist there. Even if continuous, the derivative might not exist if the graph has a sharp corner or cusp at that point. In this case, for \(x > 1\), the function simplifies to a simple constant, which is differentiable everywhere in that interval.

The function \(f(x) = (|x| - |x – 1|)^2\) can be written piecewise as:

• \(f(x) = 1\) for \(x < 0\)
• \(f(x) = (2x - 1)^2\) for \(0 \le x < 1\)
• \(f(x) = 1\) for \(x \ge 1\)

For \(x > 1\), the function is \(f(x)=1\), and its derivative is 0.