Problem 29 - Calculus Early Transcendentals

Question

a. Find the critical points of the following functions on the domain or on the given interval. b. Use a graphing utility to determine whether each critical point corresponds to a local maximum, local minimum, or neither. $$f(x)=x /\left(x^{2}+1\right)$$

Step-by-Step Solution

Answer

Answer: The critical points of the function are \(x = -1\) and \(x = 1\). At \(x = -1\), the function has a local maximum, and at \(x = 1\), it has a local minimum.

Step 1: Find the derivative of the function

To find the critical points of the function, we need to find its derivative. The function is given as: $$f(x)=\frac{x}{x^{2}+1}$$ To find the derivative, we can use the quotient rule. The quotient rule states that if we have a function \(g(x)=\frac{h(x)}{k(x)}\), then its derivative \(g'(x)\) is given by: $$g'(x) = \frac{h'(x)k(x) - h(x)k'(x)}{(k(x))^2}$$ In our case, \(h(x) = x\) and \(k(x) = x^2+1\). Therefore, \(h'(x) = 1\) and \(k'(x) = 2x\). We can now find the derivative of the function \(f(x)\): $$f'(x) = \frac{(1)(x^2+1) - (x)(2x)}{(x^2+1)^2}$$

Step 2: Simplify the derivative

Now, we need to simplify the expression for the derivative: $$f'(x) = \frac{x^2 + 1 - 2x^2}{(x^2+1)^2}$$ $$f'(x) = \frac{-x^2 + 1}{(x^2+1)^2}$$

Step 3: Find the critical points

To find the critical points, we need to set the derivative equal to zero: $$f'(x) = 0$$ $$\frac{-x^2 + 1}{(x^2+1)^2}= 0$$ The fraction is equal to zero if its numerator is equal to zero, so we have: $$-x^2 + 1 = 0$$ Solving for \(x\): $$x^2 = 1$$ Hence, the critical points are \(x = -1\) and \(x = 1\).

Step 4: Analyzing the critical points using a graphing utility

Now that we have the critical points, we will use a graphing utility to determine their nature. Use a graphing tool to plot the function \(f(x) = \frac{x}{x^2+1}\) and analyze its behavior around the critical points \(x = -1\) and \(x = 1\). By observing the graph, we can determine the following: - At \(x = -1\), the function has a local maximum (It increases as \(x\) approaches from the left and decreases as \(x\) approaches from the right). - At \(x = 1\), the function has a local minimum (It decreases as \(x\) approaches from the left and increases as \(x\) approaches from the right).

Step 5: Conclusion

The critical points of the function \(f(x) = \frac{x}{x^2+1}\) are: - A local maximum at \(x = -1\) - A local minimum at \(x = 1\)

Key Concepts

Derivative of a Function

Understanding the derivative of a function is like getting to know the heartbeat of calculus. It's a measure of how a function's output changes with respect to changes in its input - essentially detailing the rate of change. When we calculate the derivative of a function, we're looking for the slope at any point along the curve of the function.

Take, for example, a car's journey; if the function itself tells us the car's position at any given time, the derivative tells us its speed at any moment. In the step by step solution provided for the exercise, the derivative of the function \(f(x) = \frac{x}{x^2 + 1}\) was the primary focus to unearth the function's critical points - the 'interesting' points where the function either reaches a peak, a valley, or flattens out.

Quotient Rule

In calculus, the quotient rule is an elegant shortcut that allows us to slice through complex fractions and find derivatives efficiently. Imagine you have two functions stacked upon each other - one in the numerator and the other in the denominator, like a fraction. The quotient rule provides a formula to differentiate this composite function without having to decompose it into more manageable pieces first.

The rule goes like this: if you have a function \(g(x) = \frac{h(x)}{k(x)}\), then the derivative \(g'(x)\) is \(\frac{h'(x)k(x) - h(x)k'(x)}{(k(x))^2}\). Applying this to the problem, we neatly unravel the derivative of \(f(x) = \frac{x}{x^2 + 1}\) to reveal the critical points with finesse.

Local Maximum and Minimum

The peaks and troughs of a function's graph, referred to as local maximums and minimums, are akin to the highest and lowest points encountered on a roller coaster ride. A local maximum is a point where the function takes on the highest value in the immediate area; no neighboring points have a higher value. Contrastingly, a local minimum is a comfortable valley where the function attains the lowest value in its neighborhood.

Identifying these points is crucial for understanding the behavior of functions. Much like detectives, mathematicians use derivatives to locate these critical points, as observed in the provided exercise. Zeroes of the derivative indicate potential maxima or minima. Then, with the help of graphing tools, we can determine the nature of these points - whether they are indeed the turning points that, convert the fabric of the function's landscape.