Problem 39 - Calculus of a Single Variable

Question

Determine whether the Mean Value Theorem can be applied to \(f\) on the closed interval \([a, b] .\) If the Mean Value Theorem can be applied, find all values of \(c\) in the open interval \((a, b)\) such that \(f^{\prime}(c)=\frac{f(b)-f(a)}{b-a}\). If the Mean Value Theorem cannot be applied, explain why not. \(f(x)=x^{3}+2 x, \quad[-1,1]\)

Step-by-Step Solution

Answer

The values of \(c\) within the interval \((-1, 1)\) are \(±1\). The Mean Value Theorem can be applied to this function on the given interval.

Step 1: Verify the conditions of MVT

Check whether the function is continuous on the closed interval \([-1, 1]\) and differentiable on the open interval \((-1, 1)\). Since \(f(x)=x^{3}+2 x\) is a polynomial function, it is both continuous and differentiable on the entire real number line. Hence, MVT can be applied.

Step 2: Calculate the derivative of the function

Compute \(f'\) by differentiating \(f(x)=x^{3}+2 x\). The derivative is \(f'(x)=3x^{2}+2\).

Step 3: Calculate the average rate of change

Calculate the average rate of change of \(f\) over the interval \([-1,1]\) which is given by \((f(b)-f(a)) / (b-a) = (f(1)-f(-1)) / (1 - (-1)) = ((1+2)- ((-1)+(-2)))/2 = 3.\

Step 4: Solve for c

Set the derivative of the function equal to the average rate of change and solve for \(x\). This gives the equation \(3x^{2}+2 = 3\). Solving this quadratic equation yields \(x = ±1\). These are the values of \(c\) in the interval \((-1, 1)\) for which \(f'(c)\) equals to the the average rate of change of \(f\) over the interval.

Key Concepts

Continuity in Calculus

Continuity is a fundamental concept in calculus that describes the behavior of functions in a way that ensures there are no abrupt changes or gaps in their values. At its core, a function is continuous at a point if you can draw its graph at that point without lifting your pencil.

For a function to be continuous on a closed interval like \[a, b\], it must meet three conditions: firstly, it must be defined at every point within the interval, including the endpoints; secondly, it must have a limit at every point within the interval, and thirdly, the value of the function at any point must be the same as its limit at that point.

When we say that a polynomial function like \(f(x) = x^3 + 2x\) is continuous, we mean that it satisfies all these continuity conditions on the entire real number line, which includes the interval from -1 to 1. This lack of interruption or gaps in a polynomial's graph means there are no points where the behavior of the function is unpredictable or undefined, making it a good candidate for applying the Mean Value Theorem.

Differentiability

Differentiability refers to a function's ability to have a derivative. In more intuitive language, if a function is differentiable, we can calculate its slope at any point within a given range. Not all continuous functions are differentiable, but all differentiable functions are continuous. A point of non-differentiability might occur where there is a sharp corner or cusp in the function's graph.

In our exercise, the function \(f(x) = x^3 + 2x\) is not only continuous but also differentiable. That's because polynomials are smooth curves without any corners or cusps, which implies that you can find a unique tangent at every point. The derivative, or the expression for the slope of the function at any point, is given by \(f'(x) = 3x^2 + 2\). This is a crucial part of applying the Mean Value Theorem, which requires the function to be differentiable in the open interval, in our case, between -1 and 1. The derivative itself is a function and shares the continuous behavior, making it a reliable measure of how the function changes at every point within our interval.

Average Rate of Change

The average rate of change of a function over an interval is a measure of how much the function's value changes on average for each unit increase in the independent variable over that interval. It is represented as the change in the function's value, divided by the change in the independent variable.

This concept is analogous to average velocity in physics, where it's the total displacement over total time. For a function \(f\), between two points \(a\) and \(b\), it is defined as \(\frac{f(b)-f(a)}{b-a}\). This value is the slope of the secant line connecting the points \(\left(a, f(a)\right)\) and \(\left(b, f(b)\right)\) on the function's graph.

For our problem, the average rate of change of the function \(f(x) = x^3 + 2x\) from \(x=-1\) to \(x=1\) turns out to be \(\frac{f(1)-f(-1)}{1 - (-1)} = \frac{3}{2} = 3\). This represents the constant rate at which the function's value changes over that interval and becomes the value we set the function's derivative equal to when applying the Mean Value Theorem to find specific points on the function that have that slope.