Problem 30 - Calculus for Biology and Medicine

Question

Show that \(f(x)=-\left|x^{2}-4\right|\) has local maxima at \(x=2\) and \(x=-2\) but \(f(x)\) is not differentiable at \(x=2\) or \(x=-2\).

Step-by-Step Solution

Answer

The function has local maxima at \(x = 2\) and \(x = -2\) but is not differentiable at these points due to abrupt slope changes.

Step 1: Define the Function

The function given is \( f(x) = -\left|x^2 - 4\right| \). This function is the negative of the absolute value of \( x^2 - 4 \), and rewriting it gives us two cases for the absolute value: \( f(x) = -(x^2 - 4) \) if \( x^2 - 4 \geq 0 \), and \( f(x) = -(4 - x^2) \) if \( x^2 - 4 < 0 \).

Step 2: Critical Points of the Function

To find potential local maxima, we consider the critical points where derivative doesn't exist or is zero. The function changes from \( 0 \) at \( x = \pm 2 \). At these points, \( f(x) \) switches from one case to another and hence, the derivative does not exist here.

Step 3: Check Nature of Function Around Critical Points

Calculate the values of \( f(x) \) near \( x = 2 \) and \( x = -2 \). For \( x = 2 \), just below 2, \( x^2-4 \) is negative, so \( f(x) = x^2-4 = -0 \). Just above 2, \( x^2-4 \) is positive and \( f(x) \) drops to \(-4\). Similarly, for \( x = -2 \), the value changes as \( x \) crosses \( -2 \). Hence, at both points, the function dips, showing there are local maxima at these points.

Step 4: Derivative Check for Non-Differentiability

The derivative of \( f(x) \) is undefined at \( x = \pm 2 \) because the function involves the absolute value which causes a cusp or corner at these points. Both functions \(-(x^2-4)\) and \(-(4-x^2)\) have slopes that change abruptly at these points, confirming non-differentiation.

Key Concepts

Non-Differentiability

Non-differentiability can occur for a variety of reasons, particularly at points where there is a sharp turn or cusp in the graph of the function. In the given exercise, the function is defined as \( f(x) = -\left| x^2 - 4 \right| \). The presence of the absolute value function often hints at points of non-differentiability because it can create sharp
turns in the graph.
- A function is non-differentiable at a point if the slope from the left-hand side doesn't match the slope from the right-hand side.- As mentioned in the solution, at \( x=2 \) and \( x=-2 \), the function transitions between two forms, creating corners.- This mismatch at the crossover points means the derivative does not exist; thus, we refer to these as points of non-differentiability.
In our context, the derivation changes abruptly, which is the hallmark of non-differentiability at these points.

Critical Points

Critical points are vital when identifying potential local maxima or minima since these are the locations where the function's slope is zero or undefined. We find these points by attempting to take the derivative of the function and solving for zero or points where the derivative does not exist.
- A critical point occurs when \( f'(x) = 0 \) or when \( f'(x) \) is undefined.- In \( f(x) = -\left| x^2 - 4 \right| \), the critical points occur at \( x=2 \) and \( x=-2 \) because the derivative \( f'(x) \) is undefined due to the non-differentiability.
Thus, these points, \( x=2 \) and \( x=-2 \), are where potential local maxima exist, given the nature of their surrounding values.

Absolute Value Function

The absolute value function plays a key role in altering the properties of a given mathematical expression by focusing on magnitude, making all outputs non-negative. This functional form is pivotal in rendering parts of the function non-differentiable and creating critical points.
- The function \( g(x) = |x| \) provides a composite effect by being positive regardless of the input sign.- In \( f(x) = -\left| x^2 - 4 \right| \), the use of absolute value results in a piecewise function that splits at critical values \( x = \pm 2 \).- For any part where \( x^2 - 4 \geq 0 \), \( -\left| x^2 - 4 \right| \) mimics \( -(x^2-4) \), and for \( x^2 - 4 < 0 \), it resembles \( -(4-x^2) \).
The absolute value transforms \( x^2 - 4 \) into a function that pivots at critical points, molding the function into the form where non-differentiability occurs.