Question

a. Find the critical points of \(f\) on the given interval. b. Determine the absolute extreme values of \(f\) on the given interval when they exist. c. Use a graphing utility to confirm your conclusions. $$f(x)=x /\left(x^{2}+3\right)^{2} \text { on }[-2,2]$$

Short answer

Answer: The absolute minimum value is \(-\frac{1}{10}\) and occurs at \(x = -\frac{1}{\sqrt{3}}\), while the absolute maximum value is \(\frac{1}{10}\) and occurs at \(x = \frac{1}{\sqrt{3}}\).

Step-by-step solution

a. Find the critical points of \(f(x)\) on the given interval \(

\) To find the critical points of \(f(x)\), first, we must find the first derivative \(f'(x)\) of the function. Then, we will set \(f'(x) = 0\) and find the solutions that meet the given interval. $$f(x)=\frac{x}{(x^{2}+3)^{2}}$$ To find \(f'(x)\), we'll use the Quotient Rule: \((\frac{u}{v})' = \frac{u'v - uv'}{v^{2}}\). In this case, we have:$$u = x$$ $$v = (x^{2} + 3)^{2}$$Calculate the derivatives:$$u' = \frac{d}{dx}(x) = 1$$$$v' = \frac{d}{dx}((x^{2} + 3)^{2}) = 2(x^{2} + 3)(2x) = 4x(x^{2} + 3)$$Now, apply the Quotient Rule:$$f'(x) = \frac{1(x^{2} + 3)^{2} - x(4x(x^{2} + 3))}{(x^{2} + 3)^{4}}$$Simplify the expression:$$f'(x) = \frac{(x^{2} + 3)^{2} - 4x^{2}(x^{2} + 3)}{(x^{2} + 3)^{4}}$$Factor out \((x^{2} + 3)\) from the numerator:$$f'(x) = \frac{(x^{2} + 3)[(x^{2} + 3) - 4x^{2}]}{(x^{2} + 3)^{4}}$$Simplify further:$$f'(x) = \frac{(x^{2} + 3)(1 - 3x^{2})}{(x^{2} + 3)^{3}}$$Now, set \(f'(x) = 0\):$$0 = \frac{(x^{2} + 3)(1 - 3x^{2})}{(x^{2} + 3)^{3}}$$Since \((x^{2} + 3)^{3}\) is always positive, we only need to find the zeros of the numerator:$$0 = (x^{2} + 3)(1 - 3x^{2})$$Find the factors:$$x^{2} = -3$$and$$1 - 3x^{2} = 0$$Since the first equation has no real solutions, we'll only focus on the second:$$1 - 3x^{2} = 0 \Rightarrow x = \pm\frac{1}{\sqrt{3}}$$Therefore, the critical points are \(x = \frac{1}{\sqrt{3}}\) and \(x = -\frac{1}{\sqrt{3}}\), which are in the interval \([-2, 2]\).

b. Determine the absolute extreme values of \(f(x)\) on the given interval when they exist.

$ Now we will evaluate the function \(f(x)\) at the endpoints of the interval and the critical points to find the absolute maximum and minimum values. We have:$$f(-2) = \frac{-2}{(-2)^{2} + 3)^{2} = -\frac{2}{13^{2}}$$$$f(2) = \frac{2}{(2^{2} + 3)^{2}} = \frac{2}{13^{2}}$$$$f\left(\frac{1}{\sqrt{3}}\right) = \frac{1}{(1/3 + 3)^{2}} = \frac{1}{10}$$$$f\left(-\frac{1}{\sqrt{3}}\right) = -\frac{1}{(1/3 + 3)^{2}} = -\frac{1}{10}$$Comparing these values, we determine that the absolute minimum value is \(-\frac{1}{10}\) and occurs at \(x = -\frac{1}{\sqrt{3}}\), while the absolute maximum value is \(\frac{1}{10}\) and occurs at \(x = \frac{1}{\sqrt{3}}\).

c. Use a graphing utility to confirm your conclusions.

$ For this step, use a graphing utility (such as Desmos or any other graphing calculator) to graph the function \(f(x) = \frac{x}{(x^{2} + 3)^{2}}\) on the interval \([-2, 2]\). You should observe that the graph confirms the results obtained in parts a and b, where the absolute minimum and maximum values occur at \(x = -\frac{1}{\sqrt{3}}\) and \(x = \frac{1}{\sqrt{3}}\), respectively.

Key concepts

Quotient Rule

Understanding the Quotient Rule is essential for students tackling differentiation problems involving division of functions. It's a method to find the derivative of a function that is the ratio of two differentiable functions. The formula for the Quotient Rule is: \[ \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} \] where \( u \) is the numerator function and \( v \) is the denominator function.

When applying the Quotient Rule, it's important to first differentiate the numerator and the denominator separately, as these become \( u' \) and \( v' \) in our formula. Remember, you should not cancel out terms before applying the Quotient Rule, as you may lose critical information, such as potential zeros of the derivative, which could indicate critical points.

Applying the Quotient Rule successfully usually involves simplification after differentiation. Identify common factors, look for opportunities to factorise or cancel out terms, and simplify the expression to its most basic form to find the critical points or analyze the function's behavior.

First Derivative

The first derivative of a function at a point is the slope of the tangent line to the function's graph at that point. It is also the rate at which the function's value is changing at that particular point. In calculus, finding the first derivative is a fundamental skill, often done to identify the function's increasing or decreasing behavior, locate potential local extreme values, or even determine the concavity of the graph.

The process of differentiation to find the first derivative can utilize several rules, including the Power Rule, the Product Rule, the Quotient Rule, and the Chain Rule, depending on the form of the function. Finding the first derivative is also essential when we're interested in critical points because these points occur where the first derivative is either zero or does not exist. These critical points potentially hold the key to understanding the function's overall behavior and finding absolute extreme values.

Absolute Extreme Values

Absolute extreme values, or absolute extrema, refer to the highest or lowest values that a function achieves on a particular interval. These are crucial in understanding the span of function's behavior within a given range. To find absolute extrema, we typically examine the function's critical points within the interval, which we obtain by setting the first derivative equal to zero or identifying where it does not exist, in addition to evaluating the function at the endpoints of the interval.

The process of comparing these values can be summarized as follows:

1. Evaluate the function at critical points.
2. Evaluate the function at the endpoints of the interval.
3. Compare all these values to determine the smallest and largest ones.

It's vital to evaluate both the critical points and the endpoints because extreme values can occur at either of these places. The absolute maximum is the highest of these values, while the absolute minimum is the lowest.

Function Evaluation

Function evaluation is the process of determining the output of a function for a particular input. This process is pivotal in finding absolute extreme values when we have established potential critical points or endpoints of an interval. Evaluation is straightforward: you simply replace the input variable in the function with the actual values.

For instance, if we have a function given by \( f(x) = \frac{x}{(x^2 + 3)^2} \), and we want to evaluate it at \( x = \frac{1}{\text{\sqrt{3}}} \), we plug in the value in place of \( x \) in the expression and simplify.

• Ensure you follow the order of operations: Parentheses, Exponents, Multiplication and Division (from left to right), Addition and Subtraction (from left to right).
• Keep an eye out for any simplification before plugging in, which may make the evaluation easier.
• Carefully consider cases where the function might not be defined to avoid allowable errors in computation.

Correct function evaluation requires attention to detail and often a basic understanding of algebraic manipulation.