Problem 41 - Calculus an Applied Approach

Question

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 given function \(f(x) = \frac{x}{x^{2}+3}\) has its second derivative equal to zero when \(x=0\).

Step 1: Find the first derivative

Apply the quotient rule (d/dx [u/v] = (vu'-uv')/v^2), where \(u=x\) and \(v=x^2+3\). So, \(f'(x)= \frac{(1)*(x^2+3) - x(2x)}{(x^2+3)^2} = \frac{x^2+3-2x^2}{(x^2+3)^2}\). This simplifies to \(f'(x)= \frac{-x^2+3}{(x^2+3)^2}\).

Step 2: Find the second derivative

Once again apply the quotient rule for \(f''(x)\), where now \(u=-x^2+3\) and \(v=(x^2+3)^2\). This yields \(f''(x)= \frac{(2x)(x^2+3)^2 - (-2x+0)(-x^2+3)}{(x^2+3)^4}\). This simplifies to \(f''(x)= \frac{2x^3+6x}{(x^2+3)^4}\).

Step 3: Solve \(f''(x)=0\)

Set the second derivative equal to zero and solve for x. This forms the equation \(\frac{2x^3+6x}{(x^2+3)^4}=0\). We find the solution by setting the numerator equal to zero, since a fraction can only be zero when the numerator is zero. This gives us \(2x^3+6x=0\). Now factoring out, we get \(2x(x^2+3)=0\). So the values of x that satisfy the equation are \(x=0\).

Key Concepts

Quotient Rule

The quotient rule is a method used in calculus to differentiate functions that are expressed as a division of two other functions. It's particularly helpful when the function you are working with is in the form \( \frac{u}{v} \), where both \( u \) and \( v \) are functions of \( x \). To use the quotient rule, apply the formula:

• \( \frac{d}{dx}\left[\frac{u}{v}\right] = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2} \)

You differentiate \( u \) and \( v \) separately, and then plug them into the formula. Ensure you apply the product rule correctly when differentiating polynomials.
In the given exercise, \( u=x \) and \( v=x^2+3 \), making it necessary to differentiate each and apply them to the formula. This allows us to determine the first and even second derivatives of the function effectively.

Critical Points

Critical points are the values for \( x \) where the derivative of a function is either zero or undefined. These points are crucial because they can indicate potential maxima, minima, or points of inflection in a function's graph.
When solving for critical points with the second derivative test, you set \( f''(x) = 0 \) and solve for \( x \). In the exercise, the second derivative was set to zero, resulting in a polynomial equation \( 2x^3+6x=0 \).
Here, we look for values of \( x \) that make the second derivative zero as these can indicate changes in concavity, signaling potential points of inflection. To find them, factor the polynomial and solve which yields \( x=0 \).

Polynomial Equation

A polynomial equation is a mathematical expression consisting of variables and coefficients, involving operations of addition, subtraction, multiplication, and positive integer exponents. Solving these equations often involves finding the values of \( x \) that make the equation equal to zero.
In the derivative process, polynomial equations often arise when applying rules like the quotient and product rules. For example, during the calculation of the second derivative in this exercise, a polynomial \( 2x^3 + 6x \) was formed and solved.
To solve it, we set the polynomial equal to zero: \( 2x(x^2+3) = 0 \), then find the values of \( x \) that would satisfy this equation. The solution simplifies to \( x = 0 \) because it is the only real number that zeros the equation. Understanding how to manipulate and solve polynomial equations is fundamental in analyzing the behavior of functions through calculus.