Question

First Derivative Test is not exhaustive Sketch the graph of a (simple) nonconstant function \(f\) that has a local maximum at \(x=1,\) with \(f^{\prime}(1)=0,\) where \(f^{\prime}\) does not change sign from positive to negative as \(x\) increases through \(1 .\) Why can't the First Derivative Test be used to classify the critical point at \(x=1\) as a local maximum? How could the test be rephrased to account for such a critical point?

Short answer

Short Answer: The function \(f(x) = x^3(x-1)\) has a local maximum at \(x=1\) with \(f'(1)=0\), and its derivative does not change its sign around \(x=1\). The First Derivative Test fails in classifying the critical point as a local maximum because it doesn't account for "corner points" or cusps. A rephrased version of the test would be: A critical point \(x=c\) is a local maximum if \(f'(x)\) changes sign from positive to negative as \(x\) increases through \(c\), or if there is an interval around \(c\) where the slope of the function changes from increasing to decreasing but \(f'(x)\) has the same sign on both sides of \(c\).

Step-by-step solution

Sketch a function meeting the given conditions

Let's consider the function \(f(x)=x^4-x^2\). We will check if this function meets the given conditions:

Compute the first derivative

Compute the first derivative \(f'(x)\) of \(f(x)\): $$f'(x) = \frac{d}{dx}(x^4-x^2) = 4x^3-2x$$

Verify if the function has a local maximum at \(x = 1\)

To check that, we must verify if \(f'(1) = 0\): $$f'(1) = 4(1)^3 - 2(1) = 4 - 2 = 2 \neq 0$$ So, the first derivative is not equal to zero at \(x=1\). However, we are looking for a function with \(f'(1) =0\). A simple way to fix this is by adding 1 to the \(x^2\) term in the function. Therefore, let $$f(x) = x^4 - (x^2 + 1)$$ Recomputing \(f'(x)\): $$f'(x) = 4x^3 - 2x$$ Now, let's check if \(f'(1) = 0\): $$f'(1) = 4(1)^3 - 2(1) = 4 - 2 = 2 \neq 0$$ We can try to add 1 to the \(x^4\) term in the function: $$f(x) = (x^4 + 1) - x^2$$ Recomputing \(f'(x)\): $$f'(x) = 4x^3 - 2x$$ Now, let's check if \(f'(1) = 0\): $$f'(1) = 4(1)^3 - 2(1) = 4 - 2 = 2 \neq 0$$ It seems we need to change our approach: let's consider the function \(f(x) = x^3(x-1)\). Recompute \(f'(x)\): $$f'(x) = 3x^2(x-1) + x^3 = x^2(3x-3) + x^3$$ Now, let's check if \(f'(1) = 0\): $$f'(1) = (1)^2(3(1)-3) + (1)^3 = 0$$ Great! We have found a function that meets the condition \(f'(1)=0\). Now let's check the sign of \(f'(x)\) around \(x=1\).

Check the sign of the first derivative around \(x=1\)

Evaluate \(f'(x)\) for \(x<1\) and \(x>1\) to determine the sign: $$f'(0.9) = 0.9^2(3(0.9)-3) + 0.9^3 > 0$$ $$f'(1.1) = 1.1^2(3(1.1)-3) + 1.1^3 > 0$$ This function has a local maximum at \(x=1\) even though \(f'(x)>0\) for \(x<1\) and \(x>1\), which violates the First Derivative Test.

Explain why the First Derivative Test can't be used and propose a rephrased test

The First Derivative Test states that if \(f'(x)\) changes sign from positive to negative when \(x\) passes through a critical point, that point is a local maximum. However, in our example, \(f'(x)\) remains positive around \(x=1\), and yet we have a local maximum. This is because the First Derivative Test looks for changes in the slope of the tangent line, but it doesn't account for "corner points" or cusps where the function bends back on itself, creating a local maximum or minimum without a change in the sign of \(f'(x)\). To account for such cases, we could rephrase the First Derivative Test as follows: A critical point \(x=c\) is a local maximum if \(f'(x)\) changes sign from positive to negative as \(x\) increases through \(c\), or if there is some interval around \(c\) where the slope of the function changes from increasing to decreasing but \(f'(x)\) has the same sign on both sides of \(c\).

Key concepts

Critical Points

In the realm of calculus, a critical point on a function's graph represents a spot where the function's derivative is zero or undefined. Critical points are essential because they are possible locations for a function to have local maxima or minima. To identify critical points, one must calculate the first derivative of the function, set it equal to zero, and solve for the variable in question.

For example, in a function like f(x) = x³(x-1), we would first determine the first derivative f'(x) and then find the values of x that make f'(x) zero. If the derivative does not exist at some point, that is also a critical point. However, not all critical points are maxima or minima, which is why further analysis with tests like the First Derivative Test or the Second Derivative Test is necessary.

Local Maximum

A local maximum refers to a point on a function's graph where the function value is greater than all other function values in some nearby vicinity. This is a key concept within calculus when analyzing the behavior of functions. A local maximum is not necessarily the absolute highest value of the function (global maximum) but is limited to a region around the point.

For instance, the function f(x) has a local maximum at x = 1 if f(1) is higher than f(x) for all points x close to 1. Determining a local maximum involves checking the first derivative, as well as potentially the second derivative, and examining the sign changes around the critical point. Yet, the First Derivative Test mandates a sign change from positive to negative as we pass through the point, which does not necessarily need to occur in the case of a cusp or a corner point, as the standard graph sketches might not always reveal.

Calculus

Calculus is a branch of mathematics focused on the study of change and motion. It provides the tools for analyzing functions and understanding their properties such as slopes, areas under curves, and the behavior of functions as values approach infinity. Calculus is divided into two main branches: differential calculus and integral calculus.

Differential calculus involves the study of derivatives and their applications. This allows us to understand the rate at which quantities change, which is vital for discovering local extrema (maxima and minima) on graphs of functions. Integral calculus, on the other hand, involves the study of integrals and is used to calculate areas, volumes, and the total accumulation of a quantity. Together, these powerful branches of calculus are applied across various fields including physics, engineering, economics, statistics, and more.

Derivative

The derivative of a function is a measure of the function's instantaneous rate of change at any given point. It's a fundamental concept in differential calculus and essentially represents the slope of the tangent line at a point on a function’s graph. Mathematically, the derivative is defined as the limit of the average rate of change of the function as the interval over which we are measuring this rate approaches zero.

For example, the derivative of f(x) = x³(x-1) is obtained by applying the power rule and product rule of differentiation, which leads to f'(x). This derivative helps to locate critical points and analyze the function's behavior around these points. In practice, the derivative can denote many real-world concepts like velocity, momentum, and gradients of scalars in vector fields, each deeply ingrained within physics and other sciences.