Problem 40 - Calculus for Biology and Medicine

Question

Suppose that \(f(x)=x^{3}\). Explain why there exists a point \(c\) in the interval \((-1,1)\) such that \(f^{\prime}(c)=1\).

Step-by-Step Solution

Answer

The Mean Value Theorem guarantees there is at least one \(c\) in \((-1, 1)\) with \(f'(c)=1\), such as \(c=\frac{1}{\sqrt{3}}\) or \(c=-\frac{1}{\sqrt{3}}\).

Step 1: Identify Requirements for a Theorem

To show that there exists a point \(c\) where \(f'(c) = 1\), we can use the Mean Value Theorem (MVT) for derivatives. This theorem relies on two conditions: the function must be continuous on the closed interval and differentiable on the open interval. Check if function \(f(x) = x^3\) meets these conditions on the interval \([-1, 1]\).

Step 2: Check Continuity and Differentiability

The function \(f(x) = x^3\) is a polynomial, which is continuous everywhere and differentiable everywhere. Hence, it is continuous on \([-1, 1]\) and differentiable on \((-1, 1)\). Both conditions for applying the Mean Value Theorem are satisfied.

Step 3: Apply the Mean Value Theorem

According to the Mean Value Theorem, there exists a point \(c\) in \((-1, 1)\) such that \[f'(c) = \frac{f(1) - f(-1)}{1 - (-1)}.\]Calculate \(f(1) = 1^3 = 1\) and \(f(-1) = (-1)^3 = -1\).

Step 4: Calculate Derivative and Evaluate MVT Expression

Calculate the derivative \(f'(x) = 3x^2\). Now use the MVT expression:\[f'(c) = \frac{1 - (-1)}{2} = \frac{2}{2} = 1.\]Thus, we need to find \(c\) such that \(3c^2 = 1\).

Step 5: Solve for \(c\)

We solve the equation \(3c^2 = 1\) for \(c\):\[c^2 = \frac{1}{3}\ \]so \[c = rac{1}{ ext{sqrt}(3)}\text{ or } c = -rac{1}{ ext{sqrt}(3)}.\]Both solutions \(c = \frac{1}{ ext{sqrt}(3)}\) and \(c = -\frac{1}{ ext{sqrt}(3)}\) are within the interval \((-1, 1)\).

Key Concepts

Continuity

Continuity is a fundamental concept in calculus and a necessary condition for the Mean Value Theorem to apply. A function is considered continuous on an interval if there are no breaks, jumps, or holes in the graph of the function over that interval. For a polynomial, like our function \(f(x) = x^3\), continuity is a natural property since polynomials are continuous everywhere on the real number line.

In the context of the Mean Value Theorem, we need to ensure that our function \(f(x)\) is continuous on the closed interval \([-1, 1]\). Since \(f(x) = x^3\) is a polynomial, it is indeed continuous over any interval, including \([-1, 1]\). This is crucial as it ensures there are no gaps in the range of values the function can take within the interval.

As a result, continuity guarantees that the function smoothly covers all values between \(f(-1)\) and \(f(1)\), paving the way for the application of the Mean Value Theorem.

Differentiability

Differentiability is the other key requirement for the Mean Value Theorem. A function is differentiable at a point if it has a defined derivative at that point, meaning there is a tangent line that smoothly approximates the function. For the theorem, this should hold over the open interval of interest.

The function \(f(x) = x^3\) is differentiable everywhere on the real number line because the derivative, \(f'(x) = 3x^2\), exists for every \(x\). This differentiability extends to the open interval \((-1, 1)\), which allows us to use the Mean Value Theorem on this interval.

Differentiability is critical because it implies the function isn't just smooth in appearance, but its rate of change is also predictable and non-disruptive. This is why for our polynomial, possessing this property on \((-1, 1)\) enables the conclusion that there exists a point \(c\) where the slope of the tangent matches the average rate of change over the interval.

Polynomials

Polynomials are among the simplest yet most important types of functions studied in calculus. They are algebraic expressions formed by sums of powers of variables, each multiplied by coefficients.

The polynomial \(f(x) = x^3\) is a cubic polynomial, which means it has terms up to the third power of \(x\). Polynomials like this are particularly nice to work with because they are both continuous and differentiable everywhere on the real number line, which is not true for all functions.

This makes them ideal candidates for applying the Mean Value Theorem since they naturally meet the theorem's prerequisites without additional work. When dealing with a polynomial, you can rely on its smooth, unbroken nature, and its ease of differentiation to explore deeper theorems in calculus without needing to verify these properties separately. Thus, in our problem, \(f(x) = x^3\), being a polynomial, simplifies the process considerably.