Problem 25 - Contemporary Precalculus

Question

In Exercises \(15-24,\) use the techniques of Examples 4 and 5 to graph the equation in a suitable square viewing window. Use your minimum finder to approximate the \(x\) -coordinates of the lowest point on the graph of \(y=x^{3}-2 x+5\) in the window with \(0 \leq x \leq 5\) and \(-3 \leq y \leq 8 .\) The correct answer is $$ x=\sqrt{\frac{2}{3}}=.816496580928 $$ How good is your approximation?

Step-by-Step Solution

Answer

Answer: The x-coordinate of the lowest point on the graph is approximately \(x \approx 0.816496580928\).

Step 1: Find the first derivative

Calculate the first derivative of the given function, \(y=x^3-2x+5\). The first derivative represents the rate of change of the function, which will help us identify the critical points where the function changes direction. $$\frac{dy}{dx} = \frac{d}{dx}(x^3-2x+5) = 3x^2-2$$

Step 2: Determine the critical points

To determine the critical points, set the first derivative equal to zero and solve for \(x\). $$3x^2-2=0$$ This quadratic equation can be solved by the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$$ Substitute \(a=3\), \(b=0\), and \(c=-2\): $$x = \frac{\pm \sqrt{-4(-2)(3)}}{2(3)} = \frac{\pm \sqrt{24}}{6} = \frac{\pm 2\sqrt{6}}{6}=\pm \frac{\sqrt{6}}{3}$$ The two critical points are \(x=-\frac{\sqrt{6}}{3}\) and \(x=\frac{\sqrt{6}}{3}\).

Step 3: Check the critical points within the given window

By checking the critical points within the given window (\(0\leq x\leq5\), \(-3\leq y\leq8\)), notice that only \(x=\frac{\sqrt{6}}{3}\) lies within this window. So, we have only one critical point to consider: $$x=\frac{\sqrt{6}}{3} \approx 0.816496580928$$

Step 4: Identify the lowest point's x-coordinate (approximation)

We now have to determine whether the critical point represents the lowest point on the graph. We can use the second derivative test to check the concavity of the function at the critical point: $$\frac{d^2y}{dx^2} = \frac{d^2}{dx^2}(x^3-2x+5)=6x$$ At the critical point \(x=\frac{\sqrt{6}}{3}\), the second derivative is: $$\frac{d^2y}{dx^2}\Big|_{x=\frac{\sqrt{6}}{3}} = 6\left(\frac{\sqrt{6}}{3}\right) = 2\sqrt{6}$$ Since the second derivative is positive at this point, the function has a minimum point at the critical point. Thus, the x-coordinate of the lowest point on the graph is \(x=\frac{\sqrt{6}}{3}\approx 0.816496580928\).

Step 5: Compare the approximation with the correct answer

The correct answer for the x-coordinate of the lowest point is \(x=\sqrt{\frac{2}{3}}=0.816496580928\), which matches the approximation we calculated above: $$x=\frac{\sqrt{6}}{3}\approx 0.816496580928$$ As a result, our approximation is accurate.

Key Concepts

Critical Points

Understanding the concept of critical points is essential in graphing polynomials and analyzing their behavior. In calculus, a critical point is a location on the graph of a function where the derivative is zero or undefined. In simpler terms, it's where the function's slope is flat or does not exist. To find these points, we calculate the first derivative of the function and set it equal to zero, then solve for the variable.

For instance, with the polynomial function provided, whose formula is \( y = x^{3} - 2x + 5 \), we find the derivative as \( \frac{dy}{dx} = 3x^{2} - 2 \). Setting this equal to zero, we solve for \( x \) to pinpoint the critical points. Critical points are significant because they can indicate the highest or lowest values on the graph, also known as local maxima and minima, or signal a flattening out of the curve, often referred to as a saddle point.

First Derivative Test

Once we have identified the critical points, the first derivative test comes into play to determine whether these points are minima, maxima, or neither. This test involves evaluating the sign of the derivative before and after each critical point. If the derivative changes from positive to negative as it passes through a critical point, the function has a local maximum there. Conversely, if it changes from negative to positive, the function has a local minimum.

Let's apply this to our polynomial \(y = x^{3} - 2x + 5\). For the critical point \(x = \frac{\text{\textl{surd}}{6}}{3}\), we examine the sign of the derivative for values just before and after this point. A positive sign before shifting to a negative would suggest a maximum, while a negative changing to positive would indicate a minimum. It turns out, in our case, the first derivative test reveals a transition suggesting a minimum at our critical point.

Second Derivative Test

The second derivative test is another method to determine the nature of a critical point. After the first derivative helps to find where the graph may have local extrema, the second derivative tells us about the shape of the graph at these points — whether it is concave up or concave down. A positive second derivative at a critical point means the graph is concave up at that point, suggesting a local minimum. On the other hand, a negative second derivative implies concave down and a local maximum.

In our polynomial's case, taking the second derivative \( \frac{d^{2}y}{dx^{2}} = 6x \), and plugging in the critical point \(x = \frac{\text{\textl{surd}}{6}}{3}\), gives us a positive value. This aligns with a concave up graph, confirming the presence of a local minimum at the critical point. For students, it's crucial to remember that while the first derivative test gives us an idea of where extrema might occur, the second derivative test can further corroborate these findings by describing the graph's concavity at those points.