Question

Absolute maxima and minima Determine the location and value of the absolute extreme values of \(f\) on the given interval, if they exist. $$f(x)=\frac{x}{\left(x^{2}+9\right)^{5}} \text { on }[-2,2]$$

Short answer

Answer: The absolute maximum value of the function is \(\frac{2}{(13)^5}\) at \(x=-2\) and \(x=2\), and the absolute minimum value is \(\frac{-1}{(10)^5}\) at \(x=-1\) and \(x=1\).

Step-by-step solution

Find the critical points

In order to find the critical points, first, find the derivative of the function \(f(x)=\frac{x}{(x^2+9)^5}\). By applying the quotient rule, we get: $$f'(x)=\frac{(x^2 + 9)^5 \cdot 1 - x \cdot 5(x^2 + 9)^4 \cdot 2x}{(x^2 + 9)^{10}}$$ Now, simplify the expression: $$f'(x)=\frac{(x^2 + 9)^4 (9 - 10x^2)}{(x^2 + 9)^{10}}$$ The derivative will be zero or undefined at the critical points. Since the denominator is always non-zero (\(x^2 + 9 > 0\) for all \(x\)), only the numerator needs to be considered. Set the numerator equal to zero and solve for \(x\): $$(x^2 + 9)^4 (9 - 10x^2) = 0$$ Solving for \(x\), we get two critical points, namely \(x=1\) and \(x=-1\).

Evaluate the function at the endpoints and critical points

Now, evaluate the function at the two critical points (\(x=1\) and \(x=-1\)) and the two endpoints (\(x=-2\) and \(x=2\)): $$f(-2)=\frac{-2}{((-2)^2+9)^5}=\frac{-2}{(13)^5}$$ $$f(-1)=\frac{-1}{((-1)^2+9)^5}=\frac{-1}{(10)^5}$$ $$f(1)=\frac{1}{((1)^2+9)^5}=\frac{1}{(10)^5}$$ $$f(2)=\frac{2}{((2)^2+9)^5}=\frac{2}{(13)^5}$$

Compare the values and find the absolute maxima and minima

Comparing the values of \(f(x)\) at the critical points and the endpoints, we see that the function has an absolute minimum at \(x=1\) and \(x=-1\) with the value \(\frac{-1}{(10)^5}\) and an absolute maximum at \(x=2\) and \(x=-2\) with the value \(\frac{2}{(13)^5}\): Absolute minimum: \(f(-1)=f(1)=\frac{-1}{(10)^5}\) Absolute maximum: \(f(-2)=f(2)=\frac{2}{(13)^5}\)

Key concepts

Critical Points

In calculus, critical points are essential for analyzing functions and their extrema. A critical point occurs where the first derivative of a function is either zero or undefined. Investigating critical points of a function aids in identifying where it may attain its highest or lowest values on a given interval.

Identifying critical points involves calculating the derivative of the function and setting it equal to zero. In our example, after finding the derivative of the function using the quotient rule, we set the numerator of the derivative equal to zero since the denominator never becomes zero. Through this process, critical points at the x-values where the first derivative is zero are found—these points can potentially be locations of absolute extrema on the domain of the function.

For the function in the exercise \( f(x)=\frac{x}{(x^2+9)^5} \) on the interval [-2, 2], critical points are found at \( x=1 \) and \( x=-1 \). These points are candidates to be either maximums or minimums within the given interval and are crucial for further analysis in determining the absolute extrema.

Quotient Rule Differentiation

The quotient rule is a technique in calculus used for differentiating expressions where one function is divided by another. It is particularly handy when finding the derivative of a ratio of two functions. The quotient rule states that for two differentiable functions \(u(x)\) and \(v(x)\), the derivative \( u/v \) is given by
\[ \(f'(x) = \frac{v(x)u'(x) - u(x)v'(x)}{[v(x)]^2}\).\]

In the context of our problem, we apply this rule to differentiate the function \(f(x)\). With \(u(x) = x\) and \(v(x) = (x^2 + 9)^5\), we find that
\[ f'(x)=\frac{(x^2 + 9)^5 \cdot 1 - x \cdot 5(x^2 + 9)^4 \cdot 2x}{(x^2 + 9)^{10}} \].

After simplifying the derivative, we can analyze the behavior of the function by locating the critical points. Students must pay close attention to simplifying the derivative correctly to isolate the critical points accurately—this is a step where a slight mistake can lead to the wrong conclusion regarding a function's behavior.

Function Evaluation

Function evaluation is the process of finding the output of a function for specific inputs. In our exercise, evaluating the function at the discovered critical points and at the endpoints of the interval is crucial to determine the absolute extrema. To evaluate \( f(x) \) at these points, one substitutes the \( x \) value into the original function to calculate its corresponding \( y \) value.

For the function \( f(x)=\frac{x}{(x^2+9)^5} \) and the points in question—\[(-2, 1), (-1, 1), (1, 1), (2, 1)\], function evaluation provides us with the following values:

• \[ f(-2)=\frac{-2}{(13)^5} \]
• \[ f(-1)=\frac{-1}{(10)^5} \]
• \[ f(1)=\frac{1}{(10)^5} \]
• \[ f(2)=\frac{2}{(13)^5} \]

By comparing these values, we can determine that the function attains its absolute minimum at both \( x=1 \) and \( x=-1 \) and its absolute maximum at both \( x=2 \) and \( x=-2 \) on the interval [-2, 2]. Function evaluation is a simple yet powerful tool in confirming the locations of extrema and understanding the overall shape and behavior of functions over specified intervals.