Problem 30 - Calculus for Biology and Medicine

Question

[This problem illustrates the fact that \(f^{\prime \prime}(c)=0\) is not a sufficient condition for an inflection point of a twice-differentiable function.] Show that the function \(f(x)=x^{4}\) has \(f^{\prime \prime}(0)=0\) but that \(f^{\prime \prime}(x)\) does not change sign at \(x=0\) and, hence, \(f(x)\) does not have an inflection point at \(x=0\).

Step-by-Step Solution

Answer

There is no inflection point at \( x = 0 \) since \( f''(x) \) does not change sign.

Step 1: Find the First Derivative

To begin, we need to find the first derivative of the function. The function is given as \( f(x) = x^4 \). The first derivative \( f'(x) \) is found by differentiating \( f(x) \) with respect to \( x \).The derivative of \( x^4 \) is \( 4x^3 \). Thus, \( f'(x) = 4x^3 \).

Step 2: Find the Second Derivative

Next, we need to find the second derivative of the function for which we will differentiate \( f'(x) \).The second derivative \( f''(x) \) is the derivative of \( 4x^3 \), which is \( 12x^2 \). Thus, \( f''(x) = 12x^2 \).

Step 3: Evaluate the Second Derivative at x=0

Evaluate \( f''(x) \) at \( x = 0 \) to find \( f''(0) \).Substituting \( x = 0 \) into \( f''(x) = 12x^2 \), we get:\[ f''(0) = 12(0)^2 = 0 \].

Step 4: Analyze the Sign Change of f''(x) Around x=0

An inflection point requires that the second derivative changes sign at that point.Let's analyze \( f''(x) = 12x^2 \). Since \( x^2 \) is always non-negative, \( f''(x) \) is always non-negative. Thus, \( f''(x) > 0 \) for \( x eq 0 \). This indicates there is no sign change as \( x \) moves through zero.

Step 5: Conclude about the Inflection Point

Since \( f''(x) = 12x^2 \) does not change sign as \( x \) moves through zero, \(f(x) = x^4\) does not have an inflection point at \( x = 0 \), despite \( f''(0) = 0 \).

Key Concepts

Second Derivative

When we talk about the second derivative of a function, we are looking for the derivative of the first derivative. This gives us information about the curvature of the function. If you're given a function, like in the original exercise, say, \( f(x) = x^4 \), the second derivative is found by differentiating the first derivative \( f'(x) \). For \( f(x) = x^4 \), first, differentiate to find \( f'(x) = 4x^3 \). Then, differentiate \( 4x^3 \) to get \( f''(x) = 12x^2 \).

In general, the second derivative tells us if the function is concave up or concave down:

• If \( f''(x) > 0 \), the function is concave up (like a smiley face). This means it is curving upwards.
• If \( f''(x) < 0 \), the function is concave down (like a frown). This means it is curving downwards.

Determining these aspects helps us understand how the graph moves and bends. However, just because the second derivative at a point is zero does not always mean there is an inflection point there. This can sometimes be misleading, as seen in this exercise.

Sign Change

The concept of 'sign change' in the context of a second derivative deals with how the value of \( f''(x) \) behaves as you move through a particular point. Let's break it down with our provided function, \( f(x) = x^4 \).

For an inflection point, we need the second derivative \( f''(x) \) to change sign at that location. However, for \( f(x) = x^4 \), the second derivative is \( f''(x) = 12x^2 \). A look at this tells us two things:

• \( 12x^2 \) is non-negative for all \( x \).
• Since \( x^2 \) is never negative, \( 12x^2 \) never becomes negative regardless of the value of \( x \).

As a result, \( f''(x) \) never changes sign as \( x \) passes through zero. Therefore, despite \( f''(0) = 0 \), there is no sign change from negative to positive or vice versa, so there's no inflection point here. Understanding that "sign change" requires checking not just \( f''(x) = 0 \) but the behavior of signs across the point is crucial for pinpointing inflection points.

Twice-Differentiable Function

A twice-differentiable function is one for which you can take the first derivative and, importantly, the second derivative at all points in its domain. This implies that the function is not only smooth but doesn't have breaks, sharp corners, or cusps.

In the case of our function, \( f(x) = x^4 \), it is clearly differentiable once, giving us \( f'(x) = 4x^3 \). Computing further, it results in \( f''(x) = 12x^2 \), hence it is twice differentiable everywhere.

Twice-differentiable functions are crucial when discussing inflection points because the second derivative provides insight into the rates of change of the slope of the tangent lines. Just like not all functions are twice-differentiable, not all twice-differentiable points are inflection points. A point of inflection is only possible where the function is twice-differentiable and the second derivative changes sign at that point. For \( f(x) = x^4 \), despite being twice-differentiable at every point, it does not have an inflection point at \( x = 0 \) since the sign condition is not satisfied.