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. $$ f(x)=|x| ;[1,2] $$

Short answer

The MVT applies, and any \(c\) in \((1, 2)\) satisfies it.

Step-by-step solution

Verify Continuity

The Mean Value Theorem (MVT) requires the function to be continuous on the closed interval \([a, b]\). The function \(f(x) = |x|\) is continuous everywhere, including on the interval \([1, 2]\).

Verify Differentiability

The MVT also requires that the function be differentiable on the open interval \((a, b)\). For \(f(x) = |x|\), when restricted to \(x \geq 0\), its derivative is \(f'(x) = 1\). Hence, \(f(x)\) is differentiable on \((1, 2)\).

Apply the Mean Value Theorem

Since \(f(x)\) meets the conditions of the MVT, find \(c\) in the open interval \((1, 2)\) such that \[ f'(c) = \frac{f(b) - f(a)}{b - a} \] For \([1, 2]\), calculate \[ \frac{f(2) - f(1)}{2 - 1} = \frac{2 - 1}{1} = 1 \]So, set \(f'(c) = 1\).

Determine the Value of c

Since \(f'(x) = 1\) for \(x \geq 0\) and in the interval \( (1, 2) \), every \(c\) in \((1, 2)\) satisfies \(f'(c) = 1\). Therefore, every point in the interval is a potential \(c\).

Conclusion

The Mean Value Theorem applies, and all points \(c\) in the interval \((1, 2)\) satisfy the theorem's condition.

Key concepts

Continuity

When exploring the Mean Value Theorem (MVT), continuity is a crucial concept. Continuity of a function means that there are no sudden jumps or breaks in its graph. In simpler terms, you can draw the graph without lifting your pencil. For the absolute value function, denoted as \( f(x) = |x| \), the function is continuous everywhere on the real number line.
The MVT requires the function to be continuous on the closed interval \( [a, b] \). For the given exercise, we need to check if \( f(x) \) is continuous in \( [1, 2] \). Since the absolute value function is known to be continuous over all real numbers, it is of course continuous in our specified interval as well. This continuity ensures part of the requirement for the MVT is fulfilled. Without a continuous function on \([1, 2]\), the MVT could not be applied.

Differentiability

Differentiability is another criterion for using the Mean Value Theorem. A function is differentiable at a point if it has a well-defined tangent there, implying no sharp edges or cusps. In mathematical terms, it means the derivative must exist at each point.
Looking at our function, \( f(x) = |x| \), we know it has a sharp turn at \( x = 0 \). Thus, \( f(x) \) is not differentiable everywhere. However, since we are looking at the interval \([1, 2]\), which is entirely within positive \( x \) values, the absolute value function behaves like \( f(x) = x \).
In this interval, the derivative is constant, \( f'(x) = 1 \), which means \( f(x) \) is differentiable everywhere on \((1, 2)\). Hence, this criterion of the MVT is also fulfilled, so we can proceed to find potential \( c \) values.

Absolute Value Function

The absolute value function, \( f(x) = |x| \), is a fundamental part of this exercise. It represents the distance of a number from zero, meaning it always returns non-negative values regardless of whether \( x \) is positive or negative. This function is continuous over all real numbers, but it has a noticeable feature at \( x = 0 \), where it creates a cusp.
This cusp is why \(|x|\) is not differentiable at \( x = 0 \). In every other interval where \( x \geq 0 \), the function behaves like the linear function \( f(x) = x \). For this exercise over \([1, 2]\), the function behaves smoothly without any interruption or nondifferentiable points.
Understanding the behavior of the absolute value function is essential in applying tools like the Mean Value Theorem correctly, as it allows for identifying where the function meets the theorem's criteria and where it does not.