Problem 52 - Calculus and its applications

Question

Consider the function \(k\) given by $$k(x)=2|x+5|$$ a) For what \(x\) -value(s) is this function not differentiable? b) Evaluate \(k^{\prime}(-10), k^{\prime}(-7), k^{\prime}(-2),\) and \(k^{\prime}(0)\).

Step-by-Step Solution

Answer

The function is not differentiable at \(x = -5\).\(k'(-10) = -2\), \(k'(-7) = -2\), \(k'(-2) = 2\), \(k'(0) = 2\).

Step 1: Analyze the Absolute Value

The function is given by \(k(x)=2|x+5|\). The absolute value function \(|x+5|\) can create a cusp or corner point at \(x = -5\), which is where the function inside the absolute value becomes zero. At this x-value, the function may not be differentiable.

Step 2: Determine Non-Differentiable Point

Substitute \(x = -5\) into \(|x+5|\): \(|-5+5| = |0| = 0\). Therefore, the only point where \(k(x)\) might not be differentiable is \(x = -5\), as the absolute value function causes a sharp turn or cusp at this point.

Step 3: Differentiate the Function for the General Case

The function can be written as a piecewise function. When \(x > -5\), \(k(x) = 2(x+5)\), and when \(x < -5\), \(k(x) = 2(-x-5)\). So, its derivative \(k'(x)\) is \(k'(x) = 2\) for \(x > -5\) and \(k'(x) = -2\) for \(x < -5\). At \(x = -5\), the function is not differentiable.

Step 4: Evaluate Derivatives at Given Points

For \(k'(-10)\), since \(-10 < -5\), \(k'(-10) = -2\).For \(k'(-7)\), since \(-7 < -5\), \(k'(-7) = -2\).For \(k'(-2)\), since \(-2 > -5\), \(k'(-2) = 2\).For \(k'(0)\), since \(0 > -5\), \(k'(0) = 2\).

Key Concepts

Absolute Value Function

The absolute value function is an important concept in mathematics, especially in calculus. An absolute value function takes the form \,\(|x|\,\), which represents the distance of a number \(x\) from zero on the number line, without considering direction. This means it always outputs a non-negative value. For example, \,\(|3|\text{ equals }3\,\) and \,\(|-3|\text{ equals }3\,\). Thus, the absolute value changes both positive and negative inputs into positive outputs.

In the context of differentiability, absolute value functions introduce unique scenarios. They can create sharp turns or corners, technically referred to as "cusps," in the function's graph. These cusps occur where the expression inside the absolute value equals zero. For the function \(k(x) = 2|x+5|\), this happens at \(x = -5\). Detecting these points is crucial because at these points, the function might not be smooth enough to have a derivative. Instead of flowing smoothly like differentiable functions do, the graph experiences a distinct angle change.

Piecewise Functions

Piecewise functions help in understanding and interpreting absolute value functions. A piecewise function is one that is defined by different expressions depending on which interval the input falls into. They are the foundation of working with piecewise continuous functions, providing a way to elegantly describe functions like absolute values with potential cusps.

For example, the function \(k(x)=2|x+5|\) can be rewritten as a piecewise function with two parts:

• \(k(x) = 2(x+5)\) when \(x > -5\)
• \(k(x) = 2(-x-5)\) when \(x < -5\)

This representation helps in easily computing derivatives, because each piece of the function is simply a linear expression without an absolute value. Separating the function into these intervals allows for direct differentiation of each linear part.

Thus, for these individual intervals, the derivative is consistent and easy to find. In general, for the function above,

• when \(x > -5\), the derivative is \(k'(x) = 2\)
• when \(x < -5\), the derivative is \(k'(x) = -2\)

This division into piecewise parts not only simplifies the differentiation process but also highlights points where the function does not behave regularly, such as at \(x = -5\), the site where the cusp occurs.

Non-Differentiable Points

Not every point on a curve has a derivative, and non-differentiable points are examples of such places. These are points where a function doesn’t have a tangent, restricting the application of standard differentiation techniques.

For example, for the function \(k(x) = 2|x+5|\), the point \(x = -5\) is non-differentiable. The absolute value function \,\(|x+5|\,\) hits zero at \(x = -5\), causing a corner to appear on the graph of \(k(x)\).

It's important to understand that while linear pieces of the function have derivatives, at the point where these pieces meet, the slope of the tangent lines from either side differs, resulting in a sharp change in direction. As a result, there is no single-value slope (hence, no derivative) that can accurately describe the behavior of the function at the point of intersection.

This is one of the key distinctions in calculus, highlighting a limitation of differentiability. It’s a reminder that while calculus provides tools to understand change, it doesn’t uniformly apply across all points on all graphs. Thus, detecting and analyzing non-differentiable points helps in a deeper understanding of a function’s true complexity and behavior.