Problem 41 - Calculus for Biology and Medicine

Question

Suppose that \(f(x)=-x^{2}+2 .\) Explain why there exists a point \(c\) in the interval \((-1,2)\) such that \(f^{\prime}(c)=-1\).

Step-by-Step Solution

Answer

By the Mean Value Theorem, \(c = \frac{1}{2}\) in \((-1, 2)\) satisfies the condition \(f'(c) = -1\).

Step 1: Recognize the Differentiability of the Function

First, we must ascertain that the function \(f(x) = -x^2 + 2\) is differentiable on the interval \((-1, 2)\) because the function is a polynomial, and polynomials are differentiable everywhere on their domain.

Step 2: Calculate the Derivative of the Function

The derivative \(f'(x)\) represents the slope of the tangent line to the curve at any point \(x\). For \(f(x)=-x^2+2\), apply the power rule to take the derivative:\[f'(x) = \frac{d}{dx}(-x^2) + \frac{d}{dx}(2) = -2x.\]

Step 3: Formulate the Condition for the Derivative

We need to show there exists \(c\) in \((-1,2)\) such that \(f'(c)=-1\). According to Step 2, the condition becomes solving the equation:\[-2c = -1.\]

Step 4: Solve for the Specific Value of \(c\)

Solve the equation \(-2c = -1\) for \(c\) by dividing both sides by \(-2\):\[c = \frac{-1}{-2} = \frac{1}{2}.\]

Step 5: Confirm that \(c\) is within the Interval

Check that \(c = \frac{1}{2}\) is within the specified interval \((-1, 2)\). Since \(-1 < \frac{1}{2} < 2\), \(c = \frac{1}{2}\) indeed lies within the interval.

Step 6: Conclude Based on the Mean Value Theorem

Apply the Mean Value Theorem, which states if \(f(x)\) is continuous on \([a, b]\) and differentiable on \((a, b)\), there exists \(c\in (a, b)\) such that \(f'(c) = \frac{f(b) - f(a)}{b - a}\). Since conditions are met and \(f'(x) = -1\) at \(x = \frac{1}{2}\), the theorem confirms the existence of such a point.

Key Concepts

Differentiability

When we talk about differentiability, we're looking at whether a function has a derivative at each point in a certain interval. This is important because the Mean Value Theorem requires continuity and differentiability on a closed interval to apply.

For our function, which is a polynomial, differentiability is a given. Polynomials are nice functions because they are smooth and have no breaks, jumps, or corners in their graphs. This implies two key things:

• The function is continuous everywhere on the real line, including the interval ([-1, 2]).
• It's differentiable everywhere since the slope of its tangent line (the derivative) exists at each point.

Therefore, we can say with certainty that our given function \( f(x) = -x^2 + 2 \) is differentiable on the interval ((-1, 2)). This sets the stage for applying our theorem.

Polynomial Functions

Polynomial functions, like our example \( f(x) = -x^2 + 2 \), are fascinating. They are expressed in the form of a sum of powers of \( x \), such as constant, linear, quadratic, and so on. Let's look deeper into these characteristics:

• Simple expression: The function \( f(x) = -x^2 + 2 \) is quadratic, composed of a term \(-x^2\) which determines the shape, and a constant \(2\) which shifts it up or down.
• Differentiability: Polynomials can be differentiated easily, thanks to their smooth nature.
• Graph properties: The graph of \( -x^2 + 2 \) is a parabola opening downwards with its vertex at the topmost point, which is the maximum.

Understanding these properties helps in analyzing how the function behaves throughout its domain. This behavior paves the way for using calculus concepts to find points of interest, like the slope of the tangent line at different points.

Calculating Derivatives

The ability to calculate the derivative is crucial because it gives us the slope of the line tangent to the curve at any given point. For polynomial functions, calculating derivatives is straightforward through the power rule.

Here's how we derive the polynomial \(f(x) = -x^2 + 2\):

• Use the power rule, which states \( \frac{d}{dx} x^n = nx^{n-1} \).
• Apply this rule to \(-x^2\): Multiply the exponent \(2\) by the coefficient \(-1\) (giving \(-2\)) and decrease the exponent by 1 (resulting in \(-2x\)).
• Differentiate the constant \(2\) which is \(0\) because the slope of a constant is always zero.

Now, the derivative \(f'(x)\) becomes \(-2x\), which tells us how steep the tangent slope is at any given \(x\). Knowing the derivative allows us to identify specific slopes, such as having a slope of \(-1\) at some point within our interval, leading us to find the value of \(c\) that satisfies the condition \(f'(c) = -1\).