Problem 14 - 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. $$ g(x)=x^{5 / 3} ;[-1,1] $$

Step-by-Step Solution

Answer

The Mean Value Theorem does not apply because \(g(x)\) is not differentiable at \(x=0\).

Step 1: Check Continuity on the Interval

The first requirement for the Mean Value Theorem is that the function must be continuous on the closed interval \([-1, 1]\). The function \(g(x) = x^{5/3}\) is continuous for all real numbers, including on the interval \([-1, 1]\). Hence, \(g(x)\) is continuous on \([-1, 1]\).

Step 2: Check Differentiability on the Open Interval

Next, the Mean Value Theorem requires the function to be differentiable on the open interval \((-1, 1)\). To check this, we find the derivative of the function: \(g'(x) = \frac{5}{3}x^{2/3}\). The derivative exists for every \(xeq0\). Since \(g'(x)\) is not defined at \(x=0\), \(g(x)\) is not differentiable on the open interval \((-1, 1)\), because \(0\) is in this interval.

Step 3: Conclusion about the Mean Value Theorem Application

Since the function \(g(x)\) is not differentiable at \(x=0\) within the interval \((-1, 1)\), the Mean Value Theorem does not apply to the function \(g(x) = x^{5/3}\) on the interval \([-1, 1]\).

Step 4: Visualize Function to Confirm Conclusions

Sketch the graph of \(g(x)=x^{5/3}\) over the interval \([-1, 1]\). Notice the sharp corner or cusp at \(x=0\) in the graph, which confirms the lack of differentiability at that point, further supporting that the Mean Value Theorem does not apply.

Key Concepts

Continuous Function

In mathematics, a continuous function is one that does not have any gaps, jumps, or breaks in its graph. When assessing continuity, especially in the context of the Mean Value Theorem (MVT), we need to ensure the function behaves predictably between every point in a given interval.
For any function to be continuous on a closed interval \([-1, 1]\), it should satisfy two conditions:

• The function must be defined at every point in the interval.
• The limit of the function as it approaches any point from either side must equal the function's value at that point.

The given function, \( g(x) = x^{5/3} \), meets these conditions because it is continuous for all real numbers. This includes all points in the interval \([-1, 1]\). Thus, it passes the first criterion of the MVT, which is continuity on the closed interval.

Differentiability

Differentiability is a core requirement for the Mean Value Theorem alongside continuity. A function is differentiable at a point if it has a derivative at that point. This means the function's graph should not have sharp corners or cusps at any point within the interval.To test differentiability for the function \( g(x) = x^{5/3} \) on the interval \((-1, 1)\), we calculate its derivative: \[ g'(x) = \frac{5}{3}x^{2/3} \]. While the derivative exists everywhere except at \(x = 0\), differentiability fails there because \(x=0\) is part of our interval. This lack of differentiability at \(x=0\) can visually be confirmed by a cusp in the function’s graph at that point. Thus, \(g(x)\) is not differentiable on \((-1, 1)\), making the Mean Value Theorem inapplicable.

Closed Interval

A closed interval, denoted as \([a, b]\), is an important mathematical concept where both endpoints \(a\) and \(b\) are included in the interval. In the context of the Mean Value Theorem, the function needs to be continuous on such closed intervals to meet part of the theorem's criteria.In this exercise, the interval \([-1, 1]\) includes the endpoints \(-1\) and \(1\), ensuring that any analysis within this interval takes into account the boundaries as well. The continuity of \(g(x) = x^{5/3} \) over this closed interval \([-1, 1]\) is crucial, as it reassures that there are no sudden changes or discontinuities from \(-1\) to \(1\). However, when considering the Mean Value Theorem's requirement for differentiability, we refer specifically to the open interval \((-1, 1)\), which excludes these endpoints.