Problem 12 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

Find and classify all critical points. (a) \(f(x)=x^{3}-3 x+1\) (b) \(f(x)=x^{3}+3 x+1\)

Step-by-Step Solution

Answer

For the function (a), the critical points are at \(x=1\) (local minimum) and \(x=-1\) (local maximum). The function (b) has no real critical points.

Step 1: Derive the functions

For both (a) and (b), the first step is to take the derivative of the given function. For function (a) \(f(x)=x^{3}-3 x+1\), then \(f'(x)=3x^{2}-3\). And for (b) \(f(x)=x^{3}+3 x+1\), the derivative \(f'(x)=3x^{2}+3\).

Step 2: Find critical points

Next you should set each derivative equal to zero to find the critical points. For the function (a), you have \(3x^{2}-3=0\), which simplifies to \(x^{2}=1\). So the critical points are \(x=1\) and \(x=-1\). For the function (b), you have \(3x^{2}+3=0\), which simplifies to \(x^{2}=-1\). But there are no real solutions for this equation, hence function (b) has no critical points.

Step 3: Classify the critical points

Using the second derivative test, if the value of the second derivative at a critical point is positive, this point is a local minimum. If the value is negative, this point is a local maximum. If the value is zero, the test is inconclusive. For function (a), you have \(f''(x)=6x\), and evaluated at the critical points, you get \(f''(1)=6>0\) and \(f''(-1)=-6<0\), so \(x=1\) is a local minimum and \(x=-1\) is a local maximum.

Step 4: Summary

The function (a) \(f(x)=x^{3}-3 x+1\) has two critical points. \(x=1\) is a local minimum and \(x=-1\) is a local maximum. However, the function (b) \(f(x)=x^{3}+3 x+1\) has no real critical points.

Key Concepts

Derivative of a Function

Understanding the concept of the derivative of a function is crucial in calculus, as it measures how a function changes as its input changes. Technically, it represents the function's instantaneous rate of change at any given point.

For example, considering the functions given in our exercise, taking the derivative of a cubic function such as for (a) f(x) = x^3 - 3x + 1, we apply power rule which gives us f'(x) = 3x^2 - 3. Similarly, for (b) f(x)=x^3 + 3x + 1, the derivative is f'(x) = 3x^2 + 3. These derivatives help us understand how the function's output value changes in relation to changes in x.

When we set these derivatives to zero, we are essentially looking for points where the function's rate of change is nil, meaning the graph of the function has a horizontal tangent or a change in direction – these are potential critical points.

Second Derivative Test

The second derivative test is a convenient method to determine the nature of critical points found on a function's graph. After finding the first derivative and critical points, as we have with functions (a) and (b), we can apply the second derivative test to classify these points.

Take for example the second derivative of function (a), which is f''(x) = 6x. When we evaluate this at our critical points, x = 1 and x = -1, we get positive and negative values, respectively. This indicates that at x = 1, the function has a concave up shape, leading us to conclude it's a local minimum. Conversely, at x = -1, the function has a concave down shape, thus it's a local maximum. If the second derivative equals zero, more investigation is needed as the test is inconclusive in that case.

Classifying Critical Points

Critical points are where the function has a potential to change direction, and classifying critical points involves determining whether these points are local maxima, minima, or inflection points.

For function (a) from our exercise, after finding the critical points, we classify them using the second derivative test. Since f''(1) = 6 is greater than zero, x = 1 is a local minimum. Meanwhile, f''(-1) = -6 is less than zero, indicating that x = -1 is a local maximum.
Function (b) posed a unique situation where the derivative did not have real solutions for its critical points; hence, no classification was needed. It's important to remember that not all critical points lead to local minima or maxima; they sometimes indicate a saddle point or an inflection point where the curvature of the function changes.