Problem 4 - Calculus Early Transcendentals: Pearson New International Edition

Question

, a function is defined and a closed interval is given. Decide whether the Mean Value Theorem applies to the given function on the given interval. If it does, find all possible values of \(c ;\) if not, state the reason. In each problem, sketch the graph of the given function on the given interval. $$ g(x)=(x+1)^{3} ;[-1,1] $$

Step-by-Step Solution

Answer

The Mean Value Theorem applies, and the value of \(c\) is approximately 0.155.

Step 1: Verify conditions for Mean Value Theorem

The Mean Value Theorem (MVT) applies if a function is continuous on a closed interval \([a, b]\) and differentiable on the open interval \((a, b)\). Here, the function \(g(x) = (x+1)^3\) is a polynomial function, which is continuous and differentiable everywhere. Thus, \(g(x)\) is continuous on \([-1, 1]\) and differentiable on \((-1, 1)\).

Step 2: Apply the Mean Value Theorem

According to the MVT, if the conditions are met, there exists at least one \(c\) in the interval \((a, b)\) such that:\[g'(c) = \frac{g(b) - g(a)}{b - a}.\]Here, \(a = -1\) and \(b = 1\). Calculate \(g(-1)\) and \(g(1)\):\[g(-1) = ((-1)+1)^3 = 0^3 = 0,\]\[g(1) = (1+1)^3 = 2^3 = 8,\]Thus, the average rate of change is:\[\frac{g(1) - g(-1)}{1 - (-1)} = \frac{8 - 0}{2} = 4.\]

Step 3: Find the derivative and solve for \(c\)

Find the derivative \(g'(x)\) and equate it to 4:\[g'(x) = 3(x+1)^2.\]Set the derivative equal to 4:\[3(x+1)^2 = 4.\]Solve for \(x\):\[(x+1)^2 = \frac{4}{3},\]\[x+1 = \pm \sqrt{\frac{4}{3}},\]\[x = -1 \pm \sqrt{\frac{4}{3}}.\]

Step 4: Verify the value of \(c\) within \((-1, 1)\)

Calculate the approximate roots:\[x = -1 + \sqrt{\frac{4}{3}} \approx -1 + 1.155 = 0.155,\]\[x = -1 - \sqrt{\frac{4}{3}} \approx -1 - 1.155 = -2.155.\]The value \(x = 0.155\) lies within the interval \((-1, 1)\), while \(x = -2.155\) does not lie in this interval.

Step 5: Conclusion and Graph (Optional)

Since the function \(g(x)\) is continuous and differentiable on the interval, and we found \(c = 0.155\) within the interval that satisfies the MVT, the theorem applies. A sketch of \(g(x) = (x+1)^3\) on \([-1, 1]\) should show a cubic curve with points at \((-1, 0)\) and \((1, 8)\), indicating the passage through these values with the required slope of 4 at \(c = 0.155\).

Key Concepts

Continuous Function

A continuous function is one that has no breaks, jumps, or holes in its graph. Essentially, you can draw it without lifting your pencil. For the Mean Value Theorem (MVT) to apply, continuity on a closed interval \([a, b]\) is crucial. This means that at every point in the interval, the function behaves predictably and smoothly.

To identify if a function is continuous, simply observe if there are any sudden changes or breaks as you move from one point to another. Polynomial functions, such as the one in our exercise, \(g(x) = (x+1)^3\), are known for being continuous everywhere. This is because they are made up of operations (addition, multiplication, raising to a power) that do not create any discontinuities.

Differentiable Function

A differentiable function is one that has a defined derivative at every point in its domain. If a function is differentiable, it means the function has a tangent that is not vertical at every point in an interval. For the MVT to hold, differentiability on the open interval \(a, b\) is a key requirement.

Differentiability implies that the function's graph is smooth and has no sharp corners or cusps. Polynomial functions again come into play here; they are inherently differentiable everywhere. In the given exercise involving \(g(x) = (x+1)^3\), we confirm differentiability because it is a polynomial. Consequently, we can find the derivative, \(g'(x) = 3(x+1)^2\), which is needed to find the point \(c\) as per the theorem.

Polynomial Function

Polynomial functions are expressions that involve variables raised to whole number powers and combined using addition, subtraction, and multiplication. They are defined everywhere and are known for having smooth and continuous graphs.

The function \(g(x) = (x+1)^3\) is a cubic polynomial, which indicates it is of degree 3. Because polynomials never have breaks or sharp turns, they are continuously differentiable. This makes them perfect candidates for the Mean Value Theorem. In practical terms, recognizing a polynomial allows you to confidently check off the continuity and differentiability requirements for many calculus theorems, including the MVT.

Average Rate of Change

The average rate of change of a function over an interval gives us a measure of how much the function's output is changing, on average, as input changes. Calculated as the difference in function values divided by the difference in inputs, it is akin to finding the slope of the secant line connecting two points on the function's graph.

In the case of the exercise, we calculate the average rate of change over the interval \([-1, 1]\) for \(g(x) = (x+1)^3\). This results in a numerical value of 4, \(rac{8 - 0}{1 - (-1)} = 4\). The significance of this in the context of the MVT is vital: it represents the slope of the tangent line at some point \(c\) within the interval. This \(c\) is guaranteed by the theorem, as both continuity and differentiability conditions are satisfied, reinforcing the consistency of change within the interval.