Understanding the Mean Value Theorem and the Role of \(\theta\)

The question asks about the range of the parameter \(\theta\) in a specific formulation of the Mean Value Theorem (MVT):

\begin{equation} \frac{f(x+h)-f(x)}{h} = f'(x + \theta h) \end{equation}

Let's break down what the Mean Value Theorem states and how this formula relates to it.

The Mean Value Theorem (MVT) Principle

The standard Mean Value Theorem states that if a function \(f\) is:

• Continuous on the closed interval \([a, b]\)
• Differentiable on the open interval \((a, b)\)

Then, there exists at least one number \(c\) in the open interval \((a, b)\) such that:

\begin{equation} f'(c) = \frac{f(b)-f(a)}{b-a} \end{equation}

In simpler terms, the theorem guarantees that there's a point within the interval where the instantaneous rate of change (the derivative) equals the average rate of change over the entire interval.

Relating the MVT to the Given Formula

In the formula provided, \(\frac{f(x+h)-f(x)}{h} = f'(x + \theta h)\):

• We can identify \(a = x\) and \(b = x+h\).
• The term \(\frac{f(x+h)-f(x)}{h}\) represents the average rate of change of the function \(f\) between \(x\) and \(x+h\). This corresponds to \(\frac{f(b)-f(a)}{b-a}\).
• The Mean Value Theorem guarantees that there is a point \(c\) such that \(x < c < x+h\) (assuming \(h > 0\)) where \(f'(c)\) equals this average rate of change.
• The formula sets this point \(c\) equal to \(x + \theta h\).

Determining the Range of \(\theta\)

We know that the point \(c\) must lie strictly between \(x\) and \(x+h\). Therefore:

\begin{equation} x < c < x+h \end{equation}

Substitute \(c = x + \theta h\) into the inequality:

\begin{equation} x < x + \theta h < x+h \end{equation}

Now, we solve for \(\theta\). First, subtract \(x\) from all parts of the inequality:

\begin{equation} x - x < (x + \theta h) - x < (x+h) - x \end{equation}

\begin{equation} 0 < \theta h < h \end{equation}

To find the range for \(\theta\), we need to consider the sign of \(h\).

• If \(h > 0\): We can divide the inequality \(0 < \theta h < h\) by \(h\) without changing the direction of the inequalities. \begin{equation} \frac{0}{h} < \frac{\theta h}{h} < \frac{h}{h} \end{equation} \begin{equation} 0 < \theta < 1 \end{equation}
• If \(h < 0\): We divide the inequality \(0 < \theta h < h\) by \(h\), which reverses the direction of the inequalities. \begin{equation} \frac{0}{h} > \frac{\theta h}{h} > \frac{h}{h} \end{equation} \begin{equation} 0 > \theta > 1 \end{equation} This inequality \(0 > \theta > 1\) is impossible. However, if \(h < 0\), the interval is \([x+h, x]\), and the condition is \(x+h < c < x\). Substituting \(c = x + \theta h\) gives \(x+h < x + \theta h < x\). Subtracting \(x\) yields \(h < \theta h < 0\). Dividing by \(h\) (which is negative) reverses the inequalities: \(1 > \theta > 0\), which is the same as \(0 < \theta < 1\).

In both standard interpretations (whether \(h\) is positive or negative, the point \(c\) lies strictly between \(x\) and \(x+h\)), the value of \(\theta\) must be strictly between 0 and 1.

Conclusion

Based on the Mean Value Theorem, the parameter \(\theta\) in the expression \(f'(x + \theta h)\) must satisfy \(0 < \theta < 1\).