Problem 19 - Calculus Early Transcendentals: Pearson New International Edition

Question

In each of the Problems 1-21, 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. $$ f(x)=x+\frac{1}{x} ;[1,2] $$

Step-by-Step Solution

Answer

The Mean Value Theorem applies, and the possible value of \(c\) is \(\sqrt{2}\).

Step 1: Verify Function Continuity

The first condition for the Mean Value Theorem (MVT) is that the function must be continuous on the closed interval \([1, 2]\). Here, \(f(x) = x + \frac{1}{x}\) is continuous for \(x eq 0\). Since our interval \([1, 2]\) does not include 0, the function is continuous on it.

Step 2: Verify Differentiability

The second condition for the MVT is that the function must be differentiable on the open interval \((1, 2)\). The derivative of \(f(x)\) is \(f'(x) = 1 - \frac{1}{x^2}\), which exists for all \(x eq 0\). Hence, \(f(x)\) is differentiable on \((1, 2)\).

Step 3: Apply the Mean Value Theorem

Since both conditions are met, we can apply the MVT, which states there exists at least one \(c\) in \((1, 2)\) such that: \[ f'(c) = \frac{f(2) - f(1)}{2 - 1} \] We first calculate \(f(2)\) and \(f(1)\):\[ f(2) = 2 + \frac{1}{2} = 2.5 \] \[ f(1) = 1 + \frac{1}{1} = 2 \] So, \[ \frac{f(2) - f(1)}{2 - 1} = 0.5 \] Thus, we set: \[ 1 - \frac{1}{c^2} = 0.5 \]

Step 4: Solve for c

To find \(c\), solve the equation: \[ 1 - \frac{1}{c^2} = 0.5 \] Rearrange to find: \[ \frac{1}{c^2} = 0.5 \] This implies: \[ c^2 = 2 \]Taking the positive square root (since \(c\) should be between 1 and 2), we get: \[ c = \sqrt{2} \]

Step 5: Conclusion

Since \(c = \sqrt{2}\) satisfies the conditions of the MVT, it is a valid solution. Thus, the MVT applies to \(f(x)\) on the interval \([1, 2]\) and \(c = \sqrt{2}\) is the value where the theorem holds.

Key Concepts

Function Continuity

To apply the Mean Value Theorem (MVT), the first requirement is that the function must be continuous on a closed interval. For our exercise, the function given is \( f(x) = x + \frac{1}{x} \).
This function is continuous for all \( x eq 0 \).
In our specific case, the closed interval is \([1, 2]\), which does not include zero.
Thus, because there are no discontinuities within this interval, we can confidently state that the function is continuous.
Understanding continuity is crucial, as it ensures no abrupt changes or jumps in the function's behavior, which is foundational for applying the Mean Value Theorem.

Differentiability

Differentiability is the second key condition for using the Mean Value Theorem. This concept means the function has a derivative at every point within the interval.
For the given function, the derivative is:\[ f'(x) = 1 - \frac{1}{x^2} \]This derivative is defined for all \(x eq 0\).
When focusing on the open interval \((1, 2)\), we see that \(x\) remains strictly positive and does not include zero, ensuring the derivative is well-defined.
Differentiability across this interval means the function has a tangent, not a sharp cusp or break, enhancing our ability and confidence to use the Mean Value Theorem effectively.

Closed Interval

The concept of a closed interval is represented with square brackets, like \([1, 2]\).
This interval includes the endpoints, meaning the values 1 and 2 are part of the interval.
Closed intervals are essential when applying the Mean Value Theorem because continuity must hold across and up to the endpoint. In our exercise, the function maintains continuity from \(x = 1 \) to \(x = 2\).
This characteristic ensures there are no gaps or undefined points at these boundaries, fulfilling one of the critical conditions for the theorem.
In simpler terms, think of a closed interval like a stretch of highway, where you can drive from the start to the end without interruption.

Open Interval

An open interval is denoted by parentheses, such as \((1, 2)\), which means it includes values between 1 and 2 but not the endpoints themselves.
For the Mean Value Theorem, the function must be differentiable on this open interval.
In context, this means the function's derivative must not have points of non-existence or irregularities between these values.

• For the function \(f(x) = x + \frac{1}{x}\), the derivative is \(f'(x) = 1 - \frac{1}{x^2}\).
• This derivative is well-defined in the interval \((1, 2)\), as discussed earlier.

In practical terms, an open interval analogy might be a racetrack — you can race between certain barriers but must not physically touch them. The Mean Value Theorem identifies that somewhere along this path, our function's slope mirrors the overall slope from start to finish within this range.