Problem 24 - Calculus for Biology and Medicine

Question

Determine all inflection points. $$ f(x)=\ln x+\frac{1}{x}, x>0 $$

Step-by-Step Solution

Answer

The inflection point occurs at \( x = 2 \).

Step 1: Find the First Derivative

The given function is \( f(x) = \ln x + \frac{1}{x} \). First, we need to find the first derivative \( f'(x) \). The derivative of \( \ln x \) is \( \frac{1}{x} \), and the derivative of \( \frac{1}{x} \) is \( -\frac{1}{x^2} \). Therefore, the first derivative is: \[f'(x) = \frac{1}{x} - \frac{1}{x^2}.\]

Step 2: Find the Second Derivative

Next, we find the second derivative \( f''(x) \) by differentiating \( f'(x) \). The derivative of \( \frac{1}{x} \) is \( -\frac{1}{x^2} \) and the derivative of \( -\frac{1}{x^2} \) is \( \frac{2}{x^3} \). So, the second derivative is:\[f''(x) = -\frac{1}{x^2} + \frac{2}{x^3}.\]

Step 3: Set the Second Derivative to Zero

To find the inflection points, we need to set the second derivative equal to zero and solve for \( x \): \[-\frac{1}{x^2} + \frac{2}{x^3} = 0.\]

Step 4: Solve the Equation for x

Rearrange the equation to combine terms: \[\frac{2}{x^3} = \frac{1}{x^2}.\]Multiply both sides by \( x^3 \) to clear the fractions:\[2 = x.\]

Step 5: Confirm the Inflection Point

Given that \( x = 2 \) and \( x > 0 \), we substitute \( x = 2 \) back into the second derivative to check the sign change. Calculate \( f''(2) \):\[f''(2) = -\frac{1}{4} + \frac{2}{8} = -\frac{1}{4} + \frac{1}{4} = 0.\]Determine the sign of \( f''(x) \) around \( x = 2 \):- For \( x < 2 \), choose \( x = 1 \): \( f''(1) = 1 \), which is positive.- For \( x > 2 \), choose \( x = 3 \): \( f''(3) = -\frac{1}{9} + \frac{2}{27} = -\frac{1}{9} + \frac{2}{27} = -\frac{1}{27} \), which is negative. This confirms a change in concavity, thus \( x = 2 \) is an inflection point.

Key Concepts

Second Derivative

When we talk about derivatives, the first thing that usually comes to mind is the rate of change or slope of a function. The first derivative gives us that initial insight, but the second derivative takes things a step further. By differentiating the first derivative, we obtain the second derivative, which essentially tells us about the "curvature" of the function.
Using the second derivative, we can examine whether a function is curving upwards or downwards.

• If the second derivative is positive, the function is concave up, resembling a smile.
• If the second derivative is negative, the function is concave down, like a frown.

In our exercise, to find the inflection points, we determine where the second derivative changes sign. This is crucial because an inflection point is where the function transitions between curving upwards and downwards. For instance, in the provided solution, setting the equation to zero and solving reveals that our inflection point occurs at \(x = 2\).

Calculus for Biology

Calculus provides essential tools for many fields, and biology is no exception. In biological studies, understanding how populations grow, how chemicals react, or how nutrients are absorbed often requires calculus. The concept of derivatives is used widely:

• To model population change over time, where rates of increase or decrease are essential.
• To calculate how much of a certain substance is consumed or produced per unit time.
• To model the spread of diseases by understanding rate changes in infection numbers.

Inflection points, derived through the second derivative as seen in our exercise, are particularly important in biology too. They can indicate the point of fastest growth or decline in populations and help model scenarios more accurately. This deeper insight into changes helps biologists make predictions and decisions based on mathematical models, which is a powerful application of calculus beyond its theoretical roots.

Differentiation

Differentiation is the fundamental process in calculus that allows us to understand and calculate the rate at which a function is changing. By finding the derivative of a function, we gain an understanding of how that function behaves at any given point. For instance, if we have a function describing the growth of bacteria, differentiation helps us find out how quickly that population grows at any specific moment. This is achieved through step-by-step differentiation of the function:

• The first derivative grants us the slope or rate of change of the function.
• Applying differentiation again to the first derivative gives us the second derivative, which helps identify curvature and inflection points, as in our exercise.

Using these derivatives, we can decipher a wealth of information about the function and the phenomenon it represents. This is why differentiation is a cornerstone in calculus; it gives us the ability to model and understand complex real-world systems with precision.