What f’(x) equal to when 0

Question

What f&rsquo;(x) equal to when 0 < x < 1?

Prepp Team · Accepted Answer

Finding the Derivative of an Absolute Value Function

The question asks for the derivative of the function \( f(x) = (|x| - |x – 1|)^2 \) when \( 0 < x < 1 \). To find the derivative, we first need to simplify the function \( f(x) \) within the specified interval.

Simplifying the Function for \( 0 < x < 1 \)

Let's analyze the absolute value expressions \(|x|\) and \(|x – 1|\) within the interval \( 0 < x < 1 \):

For \(|x|\): Since \( 0 < x < 1 \), \( x \) is positive. Therefore, \( |x| = x \).
For \(|x – 1|\): Since \( 0 < x < 1 \), \( x - 1 \) is negative (e.g., if \( x = 0.5 \), \( x - 1 = -0.5 \)). Therefore, \( |x – 1| = -(x – 1) = 1 – x \).

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

\( f(x) = (x – (1 – x))^2 \)

Simplify the expression inside the parentheses:

\( x – (1 – x) = x – 1 + x = 2x – 1 \)

So, for \( 0 < x < 1 \), the function simplifies to:

\( f(x) = (2x – 1)^2 \)

Calculating the Derivative \( f’(x) \)

We need to find the derivative of \( f(x) = (2x – 1)^2 \) with respect to \( x \). We can use the chain rule for differentiation. The chain rule states that if \( y = u^n \), then \( \frac{dy}{dx} = n u^{n-1} \frac{du}{dx} \).

In our case, let \( u = 2x – 1 \) and \( n = 2 \). The function is \( f(x) = u^2 \).

First, find the derivative of \( u \) with respect to \( x \):

\( \frac{du}{dx} = \frac{d}{dx}(2x – 1) = 2 \)

Now, apply the chain rule:

\( f’(x) = \frac{d}{dx}((2x – 1)^2) = 2(2x – 1)^{2-1} \times \frac{d}{dx}(2x – 1) \)

\( f’(x) = 2(2x – 1)^1 \times 2 \)

\( f’(x) = 4(2x – 1) \)

Finally, distribute the 4:

\( f’(x) = 8x – 4 \)

Conclusion

When \( 0 < x < 1 \), the derivative of \( f(x) = (|x| – |x – 1|)^2 \) is \( 8x – 4 \).

Comparing this result with the given options, we find that it matches option 4.

Revision Table: Derivative Concepts

Concept	Description	Formula Example
Absolute Value Definition	\( \|a\| = a \) if \( a \ge 0 \) \( \|a\| = -a \) if \( a < 0 \)	\( \|5\|=5, \|-3\|=3 \)
Power Rule	Derivative of \( x^n \) is \( nx^{n-1} \)	\( \frac{d}{dx}(x^3) = 3x^2 \)
Chain Rule	Derivative of \( f(g(x)) \) is \( f'(g(x)) \times g'(x) \)	\( \frac{d}{dx}((2x)^3) = 3(2x)^2 \times 2 = 24x^2 \)
Derivative of a Constant	The derivative of a constant is 0.	\( \frac{d}{dx}(5) = 0 \)

Additional Information: Derivatives and Piecewise Functions

Functions involving absolute values are often piecewise functions. To find the derivative of such a function, it's crucial to first define the function explicitly in different intervals based on where the expressions inside the absolute values change sign.

For \( f(x) = (|x| - |x – 1|)^2 \), the critical points where the absolute value expressions change behavior are \( x = 0 \) (for \(|x|\)) and \( x = 1 \) (for \(|x – 1|\)). These points divide the number line into three intervals:

\( x < 0 \): \( |x| = -x \), \( |x – 1| = -(x – 1) = 1 – x \)
\( 0 \le x \le 1 \): \( |x| = x \), \( |x – 1| = -(x – 1) = 1 – x \)
\( x > 1 \): \( |x| = x \), \( |x – 1| = x – 1 \)

The question specifically asks for the derivative in the interval \( 0 < x < 1 \). In this open interval, the function is differentiable because it simplifies to a standard polynomial \( (2x-1)^2 = 4x^2 - 4x + 1 \), which is differentiable everywhere. The derivative is \( 8x - 4 \).

Note that differentiability at the points where the absolute value expressions are zero (\(x=0\) and \(x=1\)) needs separate analysis using the definition of the derivative or by checking the left and right derivatives.