Question

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

Question: Sketch a nonconstant function f with a local maximum at $x=1$ and $f^{\prime }(1)=0$. Explain why the First Derivative Test cannot be used to classify this critical point. Suggest a way to rephrase the test to account for such a critical point. Answer: A suitable example of such a function is $f(x)=x^3-3x^2+3x$. The First Derivative Test cannot be used to classify the critical point at $x=1$ because the derivative of the function, $f^{\prime }(x)=3x^2-6x+3$, does not change sign at this point. One possible way to rephrase the test is to use the Second Derivative Test, which states that if $f^{″}(x)>0$ at a critical point, there is a local minimum, if $f^{″}(x)<0$, there is a local maximum, and if $f^{″}(x)=0$, the test is inconclusive. In this case, the second derivative, $f^{″}(x)=6x-6$, is negative at x=1, which confirms that there is a local maximum at this point.

Step-by-step solution

Sketch the given function f with a local maximum at $x=1$ and $f^{\prime }(1)=0$.

Since we are asked to sketch a simple nonconstant function, we can choose a polynomial function. A suitable example for this case could be $f(x)=x^3-3x^2+3x$. This function has a local maximum at $x=1$ which can be verified by looking at its graph. Additionally, as required, the derivative of this function at $x=1$ is zero.

Find the derivative of the function f(x).

To analyze the behavior of the function around its critical points, we need to find its derivative. Applying the power rule, we can find the derivative of the given function as follows: $$f^{\prime }(x)=\frac{d}{dx}(x^3-3x^2+3x)=3x^2-6x+3$$

Identify and analyze the critical points.

A critical point is a point in the domain of a function where the derivative is either zero or undefined. In our case, we have already known that the derivative exists and equals to zero at $x=1$. Set $f^{\prime }(x)=0$, we get: $$3x^2-6x+3=0$$ However, observe that this equation has only one solution, which is $x=1$. Thus, the given function has only one critical point at $x=1$.

Explain why the First Derivative Test cannot be used to classify the critical point and suggest a rephrasing.

The First Derivative Test states that if $f^{\prime }(x)$ changes sign from positive to negative as $x$ increases through a critical point, then there is a local maximum at that point. Conversely, if the derivative changes sign from negative to positive, then there is a local minimum. However, in our case, the derivative $f^{\prime }(x)$ doesn't change its sign, and the First Derivative Test fails. One possible way to rephrase the First Derivative Test to account for such critical points would be to check the second derivative, known as the Second Derivative Test. The Second Derivative Test states that if $f^{″}(x)>0$ at a critical point, then there is a local minimum, if $f^{″}(x)<0$, then there is a local maximum, and if $f^{″}(x)=0$, the test is inconclusive. Applying this test to our example, we first find the second derivative: $$f^{″}(x)=\frac{d^2}{d{x}^2}(x^3-3x^2+3x)=6x-6$$ At $x=1$, $f^{″}(x)=-6<0$, so there is a local maximum according to the Second Derivative Test, which confirms our initial observation.

Key concepts

Critical Points

In calculus, understanding critical points is essential when examining the behavior of functions. Critically points are where a function's derivative is zero or undefined, and they provide valuable information about the function's local extrema - that is, its high and low points.

To find the critical points of a function, we set its derivative equal to zero and solve for the variable. For instance, if we have a function described by the equation $f(x)=x^3-3x^2+3x$, its derivative would be $f^{\prime }(x)=3x^2-6x+3$. Setting $f^{\prime }(x)=0$ and solving the equation will give us the critical points. In the referenced exercise, $x=1$ is a critical point since $f^{\prime }(1)=0$.

It's important not merely to identify these points but to analyze them further to determine the function's behavior at these locations. Are they points of inflection, local maxima, or local minima? That's where the First and Second Derivative Tests come into play.

Local Maximum

When discussing extremum points, a local maximum is particularly noteworthy. This point on a graph represents the highest value of a function within a certain interval. A function $f(x)$ has a local maximum at $x=c$ if $f(c)$ is greater than any value immediately preceding or following $c$.

To identify a local maximum using calculus, we look for critical points where the derivative of the function transitions from positive to negative. This implies that the slope of the tangent line to the curve moves from an upward to a downward trajectory as it crosses the critical point. However, the key here is the change in sign of the derivative. If the sign does not change, as in our exercise example at $x=1$, then the First Derivative Test cannot confirm the presence of a local maximum. Hence, another tool - the Second Derivative Test - is required to obtain conclusive evidence of the local maximum.

Second Derivative Test

The Second Derivative Test is surely a calculus student's ally when the First Derivative Test falls short. This further analysis allows us to determine the concavity of the function at a critical point, which we can then use to infer whether the point is a local maximum or minimum.

By taking the second derivative of the function $f^{″}(x)$ and evaluating it at the critical point, the results will tell us the following: if $f^{″}(x)>0$, the function is concave up, suggesting a local minimum. If $f^{″}(x)<0$, the function is concave down, which indicates a local maximum. When $f^{″}(x)=0$, the test is inconclusive, and we may need to explore further methods.

In our example, with the function $f(x)=x^3-3x^2+3x$, the second derivative at $x=1$ is $f^{″}(1)=-6$, which is less than zero. Hence, according to the Second Derivative Test, $x=1$ is indeed a local maximum. This test provides a reliable alternative when the First Derivative Test is unable to give an answer, ensuring that students can still determine the behavior of a function at critical points.