Problem 41 - College Algebra and Calculus: An Applied Approach

Question

In Exercises, find the second derivative and solve the equation \(f^{\prime \prime}(x)=0\). $$ f(x)=\frac{x}{x^{2}+3} $$

Step-by-Step Solution

Answer

The second derivative of the function \(f(x) =\frac{x}{x^{2}+3}\) is \( f''(x) = \frac{-6x}{(x^{2}+3)^{3}}\). The solution to the equation \(f''(x) = 0\) is \(x = 0\).

Step 1: Apply the quotient rule

First, the quotient rule for derivatives states that, if you have a function of the form \(h(x) = \frac{g(x)}{f(x)}\), then its derivative h'(x) can be found by: \(h'(x) = \frac{g'(x)f(x) - f'(x)g(x)}{(f(x))^2}\). Applying this rule to the function \(f(x) =\frac{x}{x^{2}+3}\), where \(g(x) = x\) and \(f(x) = x^{2}+3\), we find that \(f'(x) = \frac{(1)(x^{2}+3) - (2x)(x)}{(x^{2}+3)^{2}} = \frac{3-x^{2}}{(x^{2}+3)^{2}}\).

Step 2: Apply the quotient rule again

Next, to derive the second derivative, apply the quotient rule to \(f'(x)\). The function now looks like \(f'(x) =\frac{g(x)}{f(x)} = \frac{3-x^{2}}{(x^{2}+3)^{2}}\), where now \(g(x) = 3-x^{2}\) and \(f(x) = (x^{2}+3)^{2}\). The derivative will be: \(f''(x) = \frac{[-2x (x^{2}+3)^{2}] - [(2x)(x^{2}+3) - 2(3-x^{2})(2x)]}{(x^{2}+3)^{4}}\), which simplifies to \(f''(x) = \frac{-6x}{(x^{2}+3)^{3}}\).

Step 3: Solve the equation

Finally, solve the equation \(f''(x) = 0\). This will yield the values of x for which the function has stationary concavities. Set the function \(\frac{-6x}{(x^{2}+3)^{3}}\) equal to zero. The function will equal zero when the numerator is zero. So, -6x = 0, hence x is equal zero. Solve this equation yields a single solution,which is \(x = 0\).

Key Concepts

Quotient Rule

The Quotient Rule is a handy tool in calculus when you're tasked with finding the derivative of a ratio of two functions. Specifically, if you have a function in the form of \( h(x) = \frac{g(x)}{f(x)} \), then the derivative \( h'(x) \) is found using the formula \( h'(x) = \frac{g'(x)f(x) - g(x)f'(x)}{(f(x))^2} \). Think of it as a formula that takes into account the rate of change of both the top and the bottom part of a fraction, and how these rates of change impact the overall rate of change of the quotient. It's crucial to remember that the derivative of the numerator and the derivative of the denominator are separately calculated before they're combined according to the quotient rule formula. The correct application of this rule is imperative in reaching the solution.

Applying the Quotient Rule

Using the example \( f(x) = \frac{x}{x^2 + 3} \), you would identify \( g(x) = x \) (the numerator) and \( f(x) = x^2 + 3 \) (the denominator). Then, you'd compute \( g'(x) \) and \( f'(x) \) and plug them into the quotient rule formula. This step is essential; any error in differentiation here would lead to an incorrect result.

Derivative of a Function

The derivative of a function represents the slope of the tangent line to the function's graph at any given point. It is a fundamental concept in calculus, reflecting how the function's output value changes in response to a change in the input value. When you calculate a derivative, you're essentially looking at the limit of the function's rate of change as the interval between two points gets infinitesimally small.

Finding Derivatives

There are multiple rules to find derivatives, such as the power rule, product rule, chain rule, and, as seen in our example, the quotient rule. Each rule has a specific application according to the structure of the function. With \( f(x) = \frac{x}{x^2 + 3} \), the quotient rule is used because the function is a ratio. Notably, the process often involves simplifying complicated expressions, and mastery of algebraic techniques becomes just as important as calculus concepts. To obtain a thorough understanding, practice with a variety of functions is advisable.

Concavity of a Function

Concavity speaks to the curvature of a function's graph. Imagine a bowl: If the opening faces up, we say it's concave up; if it faces down, it's concave down. Mathematically, concavity is determined by the second derivative of a function. If the second derivative is positive over an interval, the function is concave up there. Conversely, if it's negative, the function is concave down.

Determining Concavity

Once you've found the second derivative, as in \( f''(x) = \frac{-6x}{(x^2+3)^3} \), you determine where this derivative is greater than or less than zero to find intervals of concavity. This exercise also involves solving \( f''(x) = 0 \) to find inflection points, where the function changes concavity. In the given problem, setting \( f''(x) = 0 \) led us to find \( x = 0 \) as the point at which the function's concavity changes. It is a crucial step not just for identifying the curvature of a graph but also for understanding the behavior and characteristics of the function.