Problem 4 - Contemporary Precalculus

Question

Find the coordinates of the highest or lowest point on the part of the graph of the equation in the given viewing window. Only the range of \(x\) -coordinates for the window are given \(_{i}\) you must choose an appropriate range of \(y\) -coordinates. $$y=\frac{1}{x^{2}+2 x+2} ; \quad \text { highest point when }-5 \leq x \leq 5$$

Step-by-Step Solution

Answer

Answer: The highest point is at (-1, 1).

Step 1: Find the derivative of the function

The first step is to find the derivative of the function \(y=\frac{1}{x^{2}+2 x+2}\). This can be done using the quotient rule. $$\frac{dy}{dx} = \frac{((x^{2}+2x+2)(0) - (1)(2x + 2))}{(x^{2}+2x+2)^{2}} = \frac{-(2x+2)}{(x^{2}+2x+2)^{2}}$$

Step 2: Find the critical points

To find the critical points, set the derivative equal to zero and solve for \(x\): $$\frac{-(2x+2)}{(x^{2}+2x+2)^{2}} = 0$$ Since the denominator cannot be equal to 0 for the fraction to be equal to 0, we can just solve for the numerator: $$-(2x+2) = 0$$ $$2x + 2 = 0$$ $$x=-1$$ There is only one critical point, at \(x=-1\).

Step 3: Test the endpoints and critical points

Now we need to evaluate the function at the endpoints and the critical point, \(x=-1\), to determine the highest point: At \(x = -5:\) $$y = \frac{1}{(-5)^{2}+(2(-5))+2} = \frac{1}{12}$$ At \(x = 5:\) $$y = \frac{1}{(5)^{2}+(2(5))+2} = \frac{1}{32}$$ At \(x = -1:\) $$y = \frac{1}{(-1)^{2}+2(-1)+2} = \frac{1}{1} = 1$$ The highest point is at the critical point, \(x=-1\), with a \(y\)-coordinate of \(1\).

Step 4: Choose an appropriate range of \(y\)-coordinates

Since the highest point found in the given range of \(x\)-coordinates (from \(-5\) to \(5\)) has a \(y\)-coordinate of \(1\), we can choose a reasonable range for \(y\)-coordinates considering the values of the function at the endpoints as well. A possible range for \(y\)-coordinates could be from \(0\) to \(1.5\). The highest point on the graph of the function \(y=\frac{1}{x^{2}+2 x+2}\) in the viewing window with \(-5 \leq x \leq 5\) is at the point \((-1,1)\), with an appropriate range of \(y\)-coordinates from \(0\) to \(1.5\).

Key Concepts

Critical Points in Calculus

Understanding critical points is vital in calculus, as they are where the graph of a function can change behavior. These points are found where the derivative of a function is either zero or undefined. In essence, they represent the 'turning points' where a graph can achieve its local maximum or minimum values.

To find critical points, one needs to follow a systematic approach: begin by finding the first derivative of the function since it holds the key to the function's slopes and rates of change. Once you have the derivative, set it equal to zero and solve for the variable. This process often reveals where the function plateaus or where its slope is zero. Nevertheless, remember to also check for points where the derivative is undefined as they too can be critical.

Additionally, after finding the critical points, it might be necessary to analyze them further to determine whether they are maxima, minima, or neither, which can be accomplished through various tests like the first and second derivative tests, or by comparing the values of the function at the critical points to those at the function's extremes.

Derivative of a Function

The derivative of a function is a fundamental concept in calculus that measures how the output of a function changes as its input changes. In simpler terms, it's the rate at which the function is changing at any given point. It's often represented as \( f'(x) \) for a function \( f(x) \) and calculated in various ways depending on the form of the function.

For example, to derive a simple power function, you may remember the power rule; for a product of two functions, the product rule is used. For ratios of functions, the quotient rule comes into play, which will be explored in more detail in the next section. Derivatives can be used to find critical points, calculate slopes of tangents, and solve real-world problems involving rates of change such as velocity, acceleration, and more. Its ubiquity in mathematics and related fields makes understanding how to calculate and interpret derivatives substantial.

Quotient Rule

The quotient rule is a technique used in calculus to find the derivative of a function that is the quotient of two differentiable functions. It is expressed as: If \( u(x) \) and \( v(x) \) are differentiable functions, then the derivative of the quotient \( \frac{u(x)}{v(x)} \) is given by \( \frac{u'v - uv'}{v^2} \).

This rule is particularly useful when you're dealing with fractions and need to find the rate of change. In the exercise provided, the function \( y = \frac{1}{x^{2}+2x+2} \) is an example where the numerator and denominator are both functions of \( x \) — the numerator being a constant (1) and the denominator a quadratic. Applying the quotient rule helps to find the derivative, which can then be used to determine critical points for identifying maximum values on the graph.

Remember to simplify the derivative after applying the quotient rule to make further calculations, such as setting the derivative equal to zero to find critical points, more manageable. In practical problems, understanding the quotient rule can be crucial for solving complex differential equations related to proportional relationships.