Question

In Exercises, use the derivative to identify the open intervals on which the function is increasing or decreasing. Verify your result with the graph of the function. $$ f(x)=\frac{x^{2}}{x+1} $$

Short answer

The function \( f(x)=\frac{x^{2}}{x+1} \) is always increasing for all real values of x.

Step-by-step solution

Differentiate the function

The function is a quotient, so we use the quotient rule for differentiation which states that if \[ f(x) = \frac{g(x)}{h(x)}\] then \[ f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{[h(x)]^{2}} .\] So for the function \[ f(x) = \frac{x^{2}}{x+1} ,\] the derivative \( f'(x)\) is \[ f'(x) = \frac{ (2x)(x+1) - x^{2}(1)}{(x+1)^{2}} = \frac{x}{(x+1)^{2}}.\]

Determine the critical points

Critical points occur where the derivative is equal to zero or undefined. Solving for \( f'(x) = 0,\) we get \( x=0.\) The derivative is undefined when the denominator equals zero. So, solving the equation \(x+1 = 0,\) gives \( x=-1.\) Hence, the critical points are \( x=0\) and \( x=-1.\)

Test intervals for increasing or decreasing

The critical points divide the number line into intervals: \( (-\infty, -1), (-1, 0),\) and \( (0, \infty).\) Choose a test point in each interval and evaluate the sign of the derivative at that point. If \( f'(x)\) is positive, the function is increasing. If \( f'(x)\) is negative, the function is decreasing. \n\n For the interval \( (-\infty, -1),\) choose \( x=-2, \) and find that \( f'(-2) > 0. \) Hence, the function is increasing in the interval \( (-\infty, -1).\) \n\n For the interval \( (-1, 0),\) choose \( x=-0.5, \) and find that \( f'(-0.5) > 0. \) Hence, the function is increasing in the interval \( (-1, 0). \)\n\n For the interval \( (0, \infty),\) choose \( x=1, \) and find that \( f'(1) > 0. \) Hence, the function is increasing in the interval \( (0, \infty). \)

Graph verification

Plotting \(y = f(x)\) on a graph validates that the function is increasing on the intervals found: \( (-\infty, -1), (-1, 0),\) and \( (0, \infty).\)

Key concepts

Increasing and Decreasing Functions

When we talk about whether a function is increasing or decreasing, we are referring to the behavior of the function's slopes, which are analyzed using the derivative. The derivative tells us how the function behaves at any particular point. To determine if the function is increasing or decreasing, we look at the sign of the derivative:- If the derivative, denoted as \( f'(x) \), is positive at a point, the function is increasing around that point.- If \( f'(x) \) is negative, the function is decreasing.In the exercise, the function \( f(x) = \frac{x^2}{x+1} \) required identifying its increasing and decreasing intervals. By computing the derivative \( f'(x) = \frac{x}{(x+1)^2} \), we found that the function is increasing over all intervals of \( x \), specifically noted by testing points in the intervals \( (-\infty, -1), (-1, 0), \) and \( (0, \infty) \). Thus, each test result showed a positive derivative, confirming that the function is continuously increasing over these intervals.

Testing intervals after finding critical points helps us understand the whole picture of the function’s behavior over its complete domain.

Critical Points

In calculus, critical points are where a function's derivative is either zero or undefined. These points are crucial because they can indicate where the function changes from increasing to decreasing or vice versa, or where it flattens out at a peak or trough. To find critical points, follow these steps:- Solve \( f'(x) = 0 \) to find where the slope is zero.- Check where \( f'(x) \) is undefined by analyzing the denominator of the derivative.For the function \( f(x) = \frac{x^2}{x+1} \), the derivative simplifies to \( f'(x) = \frac{x}{(x+1)^2} \). Here, the critical points occurred at \( x = 0 \) and \( x = -1 \). At \( x = 0 \), the derivative equals zero, suggesting a potential change in the slope or behavior of the function, whereas \( x = -1 \) is a location where the derivative is undefined, corresponding to a vertical asymptote in the original function.

These critical points serve as essential markers for dividing the domain into testable intervals to assess whether the function is increasing or decreasing in those regions.

Quotient Rule

The quotient rule is a technique for differentiating functions that are the ratio of two differentiable functions. This rule is particularly important when you can't simplify the expression easily to apply simpler differentiation techniques. It is expressed as:\[ f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{[h(x)]^{2}}\]Here's how it works:- \( g(x) \) is the numerator function.- \( h(x) \) is the denominator function.- Find \( g'(x) \), the derivative of the numerator.- Find \( h'(x) \), the derivative of the denominator.- Apply the formula to obtain \( f'(x) \).For the exercise at hand, where \( f(x) = \frac{x^2}{x+1} \), the quotient rule was essential. By identifying \( g(x) = x^2 \) and \( h(x) = x+1 \), and differentiating each, the expression \( f'(x) = \frac{(2x)(x+1) - x^2(1)}{(x+1)^2} \) was derived and simplified to \( \frac{x}{(x+1)^2} \).

This simplification provided a clear path to determine the behavior of \( f(x) \), such as finding critical points and evaluating the intervals for increasing or decreasing nature.