Question

The first derivative \(f^{\prime}\) is given. Find all values of \(x\) that make the function \(f(a)\) a local minimum and \((b)\) a local maximum. \(f^{\prime}(x)=x^{3}(1-x)^{2}\)

Short answer

The function \(f(x)\) has a local minimum at \(x=0\). There is no local maximum.

Step-by-step solution

Find critical points

To find critical points, set the first derivative equal to zero and solve for \(x\). This gives us the equation \(x^3(1-x)^2 = 0\). Since this expression is factored, set each factor equal to zero: \(x^3 = 0\) and \((1-x)^2 = 0\). Solving these gives the critical points at \(x = 0\) and \(x = 1\).

Determine the sign of the derivative around critical points

To determine the nature of each critical point, analyze the sign of \(f'(x)\) around each critical point. For \(x = 0\), choose points like \(-0.5, 0.5\) and for \(x = 1\), choose \(0.5, 1.5\). Calculate \(f'(x)\) at these points:- \(f'(-0.5) = (-0.5)^3(1-(-0.5))^2 < 0\)- \(f'(0.5) = (0.5)^3(1-0.5)^2 > 0\)- \(f'(1.5) = (1.5)^3(1-1.5)^2 > 0\)

Determine nature of critical points

From the sign changes calculated:- At \(x = 0\), \(f'(x)\) changes from negative to positive, indicating \(x = 0\) is a local minimum.- At \(x = 1\), \(f'(x)\) remains positive on either side, meaning there is no sign change but as polynomial root multiplicity suggest that \(f(x)\) touches the axis and could indicate a flat inflection point without a local maximum or minimum.

Conclude with local extrema

The critical point \(x = 0\) is a local minimum because \(f'(x)\) changes from negative to positive at this point. The point \(x = 1\) is not a local maximum nor minimum because there is no change in sign identified in \(f'(x)\).

Key concepts

First Derivative Test

The first derivative test is a handy tool used in calculus to identify if a critical point is a local maximum or a local minimum. When given the first derivative of a function, we initially find the critical points by setting this derivative to zero.
From there, the goal is to examine the behavior of the derivative around these critical points.

• If the derivative changes from positive to negative, the function has a local maximum at that point.
• If it changes from negative to positive, there’s a local minimum.
• If there is no change in sign, the point might be neither a maximum nor a minimum.

This makes the first derivative test an efficient method for quickly determining the nature of critical points.

Local Minimum

A local minimum is a point within a function where its value is lower than surrounding points, creating a small 'valley' in the graph. Finding these points involves using the first derivative test.
Once you have identified a critical point, as shown in the test, you check the sign of the derivative around this point.
Using the first derivative: If the derivative at a point changes from negative to positive as you cross from left to right, then you have located a local minimum. It's like descending into a valley and then starting to climb again.
Applying this to our example, the critical point at \(x = 0\) is a local minimum because \(f'(x)\) changed from negative to positive.

Local Maximum

A local maximum is the opposite of a local minimum. It's a point where a function's value is higher than nearby values, forming a 'peak' on the graph. The first derivative test helps detect these peaks by checking the sign of the derivative.
In our scenario: If the derivative switches from positive to negative when moving across a critical point, then that point is a local maximum.
It’s like climbing up to a peak and then starting to descend.
However, for the solution we evaluated, the critical point at \(x = 1\) didn’t show a sign change, indicating it wasn’t a local maximum. This highlights how sometimes critical points might not lead to a peak or trough on the graph.

Polynomial Functions

Polynomial functions are expressions that involve variables raised to whole-number exponents and can easily be graphed into smooth continuous curves. They come in various forms, such as linear, quadratic, cubic, and higher-degree polynomials.
Understanding polynomials is crucial for calculus, particularly when finding critical points and analyzing the shape of graphs.
For example, in our exercise, the function's derivative was given by the cubic polynomial \(x^3(1-x)^2\).
The nature of cubic polynomials allows the representation of more complex curves.

• The degree of the polynomial gives insight into the number of roots and how the graph behaves in terms of turning points.
• The critical points, derived from factorizing or solving the polynomial, guide us towards understanding the function's local behavior.

Cumulatively, this makes polynomial functions foundational in calculus for predicting and analyzing graph shapes.