Problem 7 - Contemporary Precalculus

Question

Find the highest point on the part of the graph of \(y=x^{3}-3 x+2\) that is shown in the given window. The answers are not all the same. (a) \(-2 \leq x \leq 0\) (b) \(-2 \leq x \leq 2\) (c) \(-2 \leq x \leq 3\)

Step-by-Step Solution

Answer

Answer: For all given windows, the highest point on the graph of the function \(y = x^3 - 3x + 2\) is at \((-1, 6)\).

Step 1: Find the first derivative of the function

First, let's find the derivative of the function \(y = x^3 - 3x + 2\). By applying the power rule, we find the first derivative: \[y' = \frac{d}{dx}(x^3 - 3x + 2) = 3x^2 - 3\]

Step 2: Find the critical points by setting the derivative to zero

Next, let's find the critical points of the function by setting the first derivative equal to zero: \[3x^2-3=0\] To solve for x, we can first divide by 3: \[x^2-1=0\] Now we can factor this equation: \[(x-1)(x+1)=0\] So, our critical points are at \(x = -1\) and \(x = 1\).

Step 3: Calculate the values of y at the critical points

Now we will find the value of y at the critical points to determine which ones are local maximums or minimums within each window. Plug the critical point values of x back into the original function: For \(x = -1\), \[y = (-1)^3 - 3(-1) + 2 = 1+3+2 = 6\] So, the point \((-1, 6)\) is a critical point of our function. For \(x = 1\), \[y = (1)^3 - 3(1) + 2 = 1-3+2 = 0\] So, the point \((1, 0)\) is a critical point of our function.

Step 4: Answer for each of the given windows

Now we will compare the critical points for each of the given windows and find the highest point in each window. (a) \(-2 \leq x \leq 0\): The only critical point in this window is at \((-1, 6)\), so that's our highest point. Answer: \((-1, 6)\). (b) \(-2 \leq x \leq 2\): Both critical points fall into this window. The point \((-1,6)\) has a higher y-value than the point \((1,0)\), so the highest point in this window is \((-1,6)\). Answer: \((-1, 6)\). (c) \(-2 \leq x \leq 3\): Both critical points fall into this window, and just like in the previous case, the point \((-1,6)\) has the highest y-value. Answer: \((-1, 6)\). In conclusion, for all given windows, the highest point on the graph of the function \(y = x^3 - 3x + 2\) is at \((-1, 6)\).

Key Concepts

Critical Points

When finding the highest point on a graph, critical points are essential. These are points where the first derivative of the function is zero or undefined, indicating a potential maximum or minimum. To identify these, we look for places on the curve where the slope is flat, signifying that the graph may be transitioning from increasing to decreasing or vice versa.

To find critical points for the function y = x^3 - 3x + 2, we calculate its first derivative and set it to zero. This gives us a set of values for x which we test within the given interval. For instance, solving 3x^2 - 3 = 0 leads to critical points at x = -1 and x = 1. These points are essential in determining the high points within different windows of x.

First Derivative Test

The First Derivative Test is a reliable method to analyze critical points and determine if they correspond to a local maximum, minimum, or neither. After finding the critical points, we examine the sign of the derivative before and after these points. If the derivative changes from positive to negative as it passes through the point, we have a local maximum. If the sign changes from negative to positive, it's a local minimum.

In the context of our function, the First Derivative Test helps us confirm that the critical point at x = -1 leads to a local maximum since the y value here is the highest when compared to other critical points. As such, for all the given windows, (-1, 6) stands out as the highest point, a conclusion supported by this test.

Power Rule

The Power Rule is a fundamental concept in calculus used to differentiate functions like y = x^3 - 3x + 2. The rule states that for any function x^n, where n is a constant, the derivative y' is nx^(n-1). It's a simple yet powerful tool that streamlines the calculation of derivatives.

For our function, applying the Power Rule yields y' = 3x^2 - 3. This rule makes it straightforward to find the first derivative, which is vital for identifying critical points and analyzing the function’s behavior. The efficiency of the Power Rule is especially evident when solving more complex equations where derivatives are the basis of critical examinations.