Understanding Differentiability of Functions

The question asks for the set of all points where the function \(f(x) = 2x|x|\) is differentiable. Differentiability means that the derivative of the function exists at that point. For a function to be differentiable at a point, it must be continuous at that point, and the left-hand derivative must equal the right-hand derivative.

Defining the Function Piecewise

The function \(f(x) = 2x|x|\) involves the absolute value function, \(|x|\). The absolute value function is defined as:

• \(|x| = x\) for \(x \ge 0\)
• \(|x| = -x\) for \(x < 0\)

Using this definition, we can write the function \(f(x)\) in a piecewise form:

• For \(x \ge 0\), \(f(x) = 2x \cdot (x) = 2x^2\).
• For \(x < 0\), \(f(x) = 2x \cdot (-x) = -2x^2\).

So, the function is:

$$ f(x) = \begin{cases} 2x^2 & \text{if } x \ge 0 \\ -2x^2 & \text{if } x < 0 \end{cases} $$

Checking Differentiability

We need to check differentiability in three intervals/points: for \(x > 0\), for \(x < 0\), and at \(x = 0\).

Case 1: Differentiability for \(x > 0\)

For \(x > 0\), the function is \(f(x) = 2x^2\). This is a polynomial function. Polynomials are differentiable for all real numbers. Therefore, \(f(x)\) is differentiable for all \(x > 0\).

The derivative for \(x > 0\) is \(f'(x) = \frac{d}{dx}(2x^2) = 4x\).

Case 2: Differentiability for \(x < 0\)

For \(x < 0\), the function is \(f(x) = -2x^2\). This is also a polynomial function. Polynomials are differentiable for all real numbers. Therefore, \(f(x)\) is differentiable for all \(x < 0\).

The derivative for \(x < 0\) is \(f'(x) = \frac{d}{dx}(-2x^2) = -4x\).

Case 3: Differentiability at \(x = 0\)

To check differentiability at \(x=0\), we need to evaluate the left-hand derivative (LHD) and the right-hand derivative (RHD) at \(x=0\). For differentiability at \(x=0\), the LHD must be equal to the RHD.

The formula for the derivative at a point \(a\) is \(f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}\). Here, \(a=0\).

First, find the function value at \(x=0\): \(f(0) = 2(0)|0| = 0\).

Left-Hand Derivative (LHD) at \(x=0\):

$$ \text{LHD} = \lim_{h \to 0^-} \frac{f(0+h) - f(0)}{h} = \lim_{h \to 0^-} \frac{f(h) - 0}{h} = \lim_{h \to 0^-} \frac{f(h)}{h} $$

Since \(h \to 0^-\), \(h\) is negative. For negative values of \(x\), \(f(x) = -2x^2\). So, \(f(h) = -2h^2\) when \(h < 0\).

$$ \text{LHD} = \lim_{h \to 0^-} \frac{-2h^2}{h} = \lim_{h \to 0^-} (-2h) = -2(0) = 0 $$

Right-Hand Derivative (RHD) at \(x=0\):

$$ \text{RHD} = \lim_{h \to 0^+} \frac{f(0+h) - f(0)}{h} = \lim_{h \to 0^+} \frac{f(h) - 0}{h} = \lim_{h \to 0^+} \frac{f(h)}{h} $$

Since \(h \to 0^+\), \(h\) is positive. For non-negative values of \(x\), \(f(x) = 2x^2\). So, \(f(h) = 2h^2\) when \(h > 0\).

$$ \text{RHD} = \lim_{h \to 0^+} \frac{2h^2}{h} = \lim_{h \to 0^+} (2h) = 2(0) = 0 $$

Since LHD at \(x=0\) (which is 0) equals RHD at \(x=0\) (which is also 0), the function \(f(x)\) is differentiable at \(x=0\).

Conclusion on Differentiability

We found that the function \(f(x)\) is differentiable:

• For all \(x > 0\)
• For all \(x < 0\)
• At \(x = 0\)

Combining these, the function \(f(x) = 2x|x|\) is differentiable for all real numbers.

The set of all real numbers is represented by the interval \((-\infty, \infty)\).

Summary of Differentiability Findings

The function is differentiable everywhere on the real number line.

Revision Table: Key Calculus Concepts

Additional Information: Differentiability and Absolute Value

Functions involving absolute values often require special attention at the points where the expression inside the absolute value becomes zero. For \(|x|\), this point is \(x=0\). For a general function \(g(x)|h(x)|\), one would investigate points where \(h(x) = 0\).

In this specific case, \(f(x) = 2x|x|\), the point of interest is \(x=0\).

Consider a simpler function like \(g(x) = |x|\). This function is continuous at \(x=0\), but not differentiable there. The LHD at 0 is -1, and the RHD at 0 is 1. Since LHD \(\ne\) RHD, \(|x|\) is not differentiable at 0.

However, when \(|x|\) is multiplied by \(x\) (or \(x^n\) for \(n \ge 1\)), the resulting function tends to become differentiable at \(x=0\). For \(f(x) = x|x|\), the LHD and RHD at 0 are both 0. For \(f(x) = x^2|x|\), the LHD and RHD at 0 are both 0. Generally, \(f(x) = x^n |x|\) is differentiable at \(x=0\) if \(n \ge 1\).

In our function \(f(x) = 2x|x|\), we have the term \(x|x|\) multiplied by a constant 2. Since \(x|x|\) is differentiable at \(x=0\), \(2x|x|\) is also differentiable at \(x=0\). The differentiability on the intervals \(x < 0\) and \(x > 0\) is straightforward as the function is a polynomial in those regions.