Problem 64 - Contemporary Precalculus

Question

Do Exercise 63 with \(f(x)=x^{8}\) in place of \(f(x)=x^{5}\)

Step-by-Step Solution

Answer

Answer: The critical point of the function \(f(x) = x^8\) is \(x = 0\), and it is a local minimum.

Step 1: Understand Exercise 63

Exercise 63 asks us to find the critical points of the given function and then determine their nature (i.e., whether the critical points are local maxima, local minima, or saddle points).

Step 2: Define the function and its derivative

Here, we are given the function \(f(x) = x^8\). To find the critical points, we need to find the first derivative of the function. The derivative of \(f(x) = x^8\) is \(f'(x) = 8x^7\).

Step 3: Find critical points

Critical points occur when the first derivative \(f'(x)\) is equal to 0 or does not exist. In this case, the first derivative always exists because it is a polynomial function. So, to find the critical point, set \(f'(x)\) equal to 0: \(8x^7 = 0 \Rightarrow x^7 = 0 \Rightarrow x = 0\). There is only one critical point at \(x = 0\).

Step 4: Determine the nature of the critical point

To determine the nature of the critical point, we can use the second derivative test. So, let's find the second derivative of the given function: \(f''(x) = \frac{d^2}{dx^2} (x^8) = \frac{d}{dx} (8x^7) = 56x^6\). Now, evaluate the second derivative at the critical point \(x = 0\): \(f''(0) = 56(0)^6 = 0\). Since the second derivative at the critical point is equal to 0, the second derivative test is inconclusive. However, we can analyze the behavior of the function around the critical point to determine its nature.

Step 5: Analyze the function around the critical point

Note that \(f(x)\) is an even power function. Therefore, as \(x\) moves away from the critical point in both the positive and negative direction, \(f(x)\) becomes positive and always increases. This indicates that \(x = 0\) is a local minimum. Final answer: The critical point of \(f(x) = x^8\) is \(x = 0\), and it is a local minimum.

Key Concepts

Function Derivatives

When studying precalculus or calculus, one of the fundamental concepts you'll encounter is the derivative of a function. Derivatives represent the rate at which a function is changing at any given point. Technically speaking, the derivative at a particular point is the slope of the tangent line to the function at that point.

For example, if you have a function like the one in our exercise, \(f(x) = x^8\), taking the derivative means finding a new function, \(f'(x)\), that gives us the rate of change of \(f(x)\). The derivative of a power function, according to the power rule, is found by multiplying the exponent by the coefficient and then subtracting one from the exponent. In our case, the derivative of \(f(x) = x^8\) is therefore \(f'(x) = 8x^7\).

Understanding how to find the derivative of a function is crucial because it allows you to determine critical points which are points on the graph where the function could change direction – leading to local maxima, minima, or saddle points.

Second Derivative Test

The second derivative test is a method used to classify the nature of critical points found in the first derivative test. You find the second derivative, \(f''(x)\), of the original function and then evaluate it at each critical point.

If the result is positive, the critical point is a local minimum, because the function curves upwards at that point. Conversely, if the result is negative, the critical point is a local maximum, given the function curves downwards. When the second derivative at the critical point is zero, the test is inconclusive; the point could be a local minimum, local maximum, or a saddle point. It is then often necessary to examine the function's behavior around the critical point to make a final determination.

In the context of our problem with \(f(x) = x^8\), the second derivative is \(f''(x) = 56x^6\). At the critical point \(x = 0\), \(f''(0)\) equals zero, rendering the second derivative test inconclusive. Additional analysis of the function's shape is required to determine the nature of the critical point.

Local Minimum

A local minimum of a function is a point where the function's value is lower than at all other nearby points. It's like being at the bottom of a small valley on the graph of the function. Discovering the local minimum is essential for various applications, including optimization problems where you need to minimize cost or maximize efficiency.

In our specific function \(f(x) = x^8\), though the second derivative test doesn't help us, examining the function shows us that for any small value of \(x\), whether positive or negative, when raised to the eighth power, the result is positive. Thus, as \(x\) moves away from zero, \(f(x)\) always increases on either side of the critical point. This kind of analysis leads to the conclusion that the critical point at \(x = 0\) is indeed a local minimum – the lowest point in its immediate vicinity.