Problem 27 - Calculus for Biology and Medicine

Question

Determine all inflection points. \(f(x)=\frac{x^{2}}{x^{2}+1}\)

Step-by-Step Solution

Answer

The inflection points are at \( x = \pm \frac{\sqrt{3}}{3} \).

Step 1: Find the First Derivative

To determine inflection points, we first need the first derivative of the function. The function given is \( f(x) = \frac{x^2}{x^2 + 1} \). We will use the quotient rule for differentiation which states that \( \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \). Here \( u = x^2 \) and \( v = x^2 + 1 \). Therefore, the derivatives are \( u' = 2x \) and \( v' = 2x \). Substituting into the quotient rule, we get:\[f'(x) = \frac{(2x)(x^2 + 1) - (x^2)(2x)}{(x^2 + 1)^2} = \frac{2x^3 + 2x - 2x^3}{(x^2 + 1)^2} = \frac{2x}{(x^2 + 1)^2}\]

Step 2: Find the Second Derivative

Next, we find the second derivative of \( f(x) \) to identify inflection points. We'll differentiate \( f'(x) = \frac{2x}{(x^2 + 1)^2} \) again using the quotient rule. Let \( u = 2x \) and \( v = (x^2 + 1)^2 \). So, \( u' = 2 \) and \( v' = 2(x^2 + 1)(2x) = 4x(x^2 + 1) \). Now, applying the quotient rule:\[f''(x) = \frac{(2)(x^2 + 1)^2 - (2x)(4x(x^2 + 1))}{((x^2 + 1)^2)^2}\]\[= \frac{2(x^2 + 1)^2 - 8x^2(x^2 + 1)}{(x^2 + 1)^4}\]Expand and simplify:The numerator becomes:\( 2(x^4 + 2x^2 + 1) - 8x^4 - 8x^2 \).Simplifying gives us:\( 2x^4 + 4x^2 + 2 - 8x^4 - 8x^2 \).Combine terms:\[ = -6x^4 - 4x^2 + 2 \].Thus,\[f''(x) = \frac{-6x^4 - 4x^2 + 2}{(x^2 + 1)^4}\]

Step 3: Set the Second Derivative Equal to Zero

Inflection points occur where the second derivative is zero or undefined. We set \( f''(x) = 0 \):\[\frac{-6x^4 - 4x^2 + 2}{(x^2 + 1)^4} = 0\]Since the denominator \((x^2 + 1)^4\) never equals zero for real values, we only need to find where the numerator equals zero:\[-6x^4 - 4x^2 + 2 = 0\]Let \( y = x^2 \); this becomes:\[-6y^2 - 4y + 2 = 0\]Multiply through by \(-1\):\[6y^2 + 4y - 2 = 0\]

Step 4: Solve the Quadratic Equation

We need to solve:\[ 6y^2 + 4y - 2 = 0 \]Use the quadratic formula: \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 6 \), \( b = 4 \), \( c = -2 \).\[y = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 6 \cdot (-2)}}{2 \cdot 6}\]\[= \frac{-4 \pm \sqrt{16 + 48}}{12}\]\[= \frac{-4 \pm \sqrt{64}}{12}\]\[= \frac{-4 \pm 8}{12}\]This gives us two solutions: \( y = \frac{4}{12} = \frac{1}{3} \) and \( y = \frac{-12}{12} = -1 \). Only \( y = \frac{1}{3} \) is valid since \( y = x^2 \) and must be non-negative.

Step 5: Determine the x-values for Inflection Points

Recalling \( y = x^2 \), for \( y = \frac{1}{3} \), we solve:\[x^2 = \frac{1}{3}\]\[x = \pm \sqrt{\frac{1}{3}} = \pm \frac{1}{\sqrt{3}} = \pm \frac{\sqrt{3}}{3}\]Thus, the inflection points are at \( x = \pm \frac{\sqrt{3}}{3} \).

Key Concepts

Derivative

The derivative of a function tells us how the function's output changes as its input changes. It's like measuring the speed of a car; if the function represents the distance, then the derivative represents the velocity.
For example, to find the derivative of a function like \(f(x) = \frac{x^2}{x^2 + 1}\), we need a rule that helps us differentiate functions that are expressed as one function divided by another. This is where the "Quotient Rule" comes into play, which helps us find derivatives for rational functions very effectively.

• Helps find slopes of functions.
• Gives insight about behaviour of functions like increase or decrease.
• Vital for optimization problems.

Quotient Rule

When you have a function that is a ratio of two other functions, the quotient rule is your friend. It helps you find the derivative of these complex looking ratios. The quotient rule states:\[\left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2}\] where \(u'\) and \(v'\) are the derivatives of \(u\) and \(v\) respectively.
For instance, if \(u = x^2\) and \(v = x^2 + 1\), then:

• Find \(u'\) and \(v'\).
• Apply the formula.
• Simplify the result.

This rule allows you to handle differentiation of quotients with precision. Just remember to carefully manage the numerator and denominator components to avoid common mistakes.

Second Derivative

The second derivative measures the rate of change of the rate of change. This means it's used to determine what happens beyond the first level of changes. It's like figuring out if a car is speeding up or slowing down after knowing its speed.
When we talk about inflection points, the second derivative comes in handy because these points appear where the second derivative equals zero.
For example, if you differentiate \(f'(x)\) = \(\frac{2x}{(x^2 + 1)^2}\) again using the quotient rule, you'll find \(f''(x)\). Solving \(f''(x) = 0\) will lead you towards discovering the function's nature.

• Helps in identifying the concavity.
• Used to find inflection points.
• Makes curves more understandable.

Quadratic Equation

A quadratic equation is an algebraic equation of the form \(ax^2 + bx + c = 0\). It typically has two solutions, which can be found using the quadratic formula:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]This formula is a staple for solving quadratic equations, no matter how complicated they might look.
For example, when simplifying the step for finding inflection points by solving \(-6y^2 - 4y + 2 = 0\), you rearrange it to \(6y^2 + 4y - 2 = 0\) before applying the quadratic formula.Applying the formula allows you to navigate through algebraic manipulations with ease, finding the roots crucial for further analysis.

• Provides methods to solve for roots.
• Useful for modeling numerous problems.
• Key in understanding parabolas and their properties.