Problem 39 - Calculus for Biology and Medicine

Question

Let \(f(x)=x(1-x)\). Use the MVT to find an interval that contains a number \(c\) such that \(f^{\prime}(c)=0\).

Step-by-Step Solution

Answer

The interval \((0, 1)\) contains the number \( c = \frac{1}{2} \) where \( f'(c) = 0 \).

Step 1: State the Mean Value Theorem (MVT)

The Mean Value Theorem states that if a function \( f \) is continuous on the closed interval \([a, b]\) and differentiable on the open interval \((a, b)\), then there exists a number \( c \) in \((a, b)\) such that \( f'(c) = \frac{f(b) - f(a)}{b - a} \).

Step 2: Define Derivative of Given Function

First, find the derivative of \( f(x) = x(1-x) \). Using the product rule \((uv)' = u'v + uv'\), where \( u = x \) and \( v = 1-x \), the derivative \( f'(x) = 1 - 2x \).

Step 3: Determine Interval for MVT

We need an interval \([a,b]\) over which \( f \) is both continuous and differentiable. Function \( f(x) = x(1-x) \) is a polynomial, therefore continuous and differentiable everywhere. We'll use the interval \([0,1]\) since it's a common interval for such problems.

Step 4: Calculate Mean Slope on Interval

Compute the slope of the secant line on \([0,1]\):\[ \frac{f(1) - f(0)}{1 - 0} = \frac{0 - 0}{1} = 0. \]

Step 5: Solve Equation for c

According to MVT, \( f'(c) = 0 \) for some \( c \) in \((0,1)\). From \( f'(x) = 1 - 2x \), set \( 1 - 2c = 0 \), solve for \( c \): \( 2c = 1 \) which gives \( c = \frac{1}{2} \).

Step 6: Verify the Interval

The solution \( c = \frac{1}{2} \) lies within the interval \((0,1)\), thus confirming that \( [0, 1] \) is an appropriate interval for \( c \) using the MVT.

Key Concepts

Understanding Calculus

Calculus is the branch of mathematics that deals with rates of change and quantities under the influence of a continuous increase or decrease. It is essential for understanding the nature of changes in any function that varies continuously.
In the context of the Mean Value Theorem (MVT), calculus helps us identify specific values within a function’s interval where instantaneous rate matches the average rate over that interval.
In the exercise, we are examining a function \( f(x) = x(1-x) \), a classic quadratic form, to utilize calculus tools in calculating its change behaviors.

The Role of Derivatives

Derivatives are a cornerstone of calculus that give us the instantaneous rate of change of a function. They indicate how a function evolves at any point, providing insights into things like velocity and acceleration when we’re talking about physical movement.
For the function \( f(x) = x(1-x) \), its derivative \( f'(x) = 1 - 2x \) is derived using the product rule, demonstrating how the function's slope changes at different points.
Understanding this derivative is crucial for applying the MVT, as the theorem states that there is a special number \( c \) in the open interval \((a, b)\) where the derivative at that point equals the average slope of the function over that interval. This principle is illustrated by setting \( f'(c) = 0 \).

Importance of Continuity

Continuity in functions signifies that there are no breaks, jumps, or points of discontinuity in the graph of the function within a particular interval. It implies that the function behaves predictably without abrupt changes.
In applying the Mean Value Theorem, continuity is a necessary precondition.
For our quadratic function \( f(x) = x(1-x) \), it is continuous (and differentiable) everywhere because it is a polynomial.
Therefore, it meets the prerequisites for the MVT application on the interval \([0, 1]\), guaranteeing that the function progresses smoothly, allowing us to find a point \( c \) where the derivative is zero. This assists us in identifying areas where changes balance out efficiently.