Question

$$ \begin{array}{l}{\text { Multiple Choice If } f(x)=\cos x, \text { then the Mean Value }} \\ {\text { Theorem guarantees that somewhere between } 0 \text { and } \pi / 3, f^{\prime}(x)=} \\ {\text { (A) }-\frac{3}{2 \pi} \quad \text { (B) }-\frac{\sqrt{3}}{2} \quad(\mathbf{C})-\frac{1}{2} \quad \text { (D) } 0}\end{array} $$

Short answer

The short answer to the exercise is (A) -\(\frac{3}{2\pi}\)

Step-by-step solution

Recall the derivative of the cosine function

The derivative of \(f(x) = \cos(x)\) is \(f'(x) = -\sin(x)\). We will use this in the next step to figure out the value of \(f'(x)\) for our particular range.

Apply the Mean Value Theorem

According to the Mean Value Theorem, there exists at least one \(c\) in the range (0, \(\pi / 3\)) such that \(f'(c) = \frac{f(b) - f(a)}{b - a}\). Here, \(a = 0\), \(b = \pi / 3\), \(f(a) = \cos(0)\) and \(f(b) = \cos(\pi / 3)\). Plugging these values into the Mean Value Theorem gives us: \(f'(c) = \frac{\cos(\pi / 3) - \cos(0)}{\pi / 3 - 0}\)

Compute the value of the derivative

Compute the value of \(f'(c)\) using the formula found in step 2: \(f'(c) = \frac{1/2 - 1}{\pi/3 - 0} = -\frac{1/2}{\pi/3} = -\frac{3}{2\pi}\)

Key concepts

Derivative of Cosine Function

Understanding the derivative of the cosine function is quite fundamental when studying calculus. The cosine function, denoted as \( f(x) = \cos(x) \), has a derivative that represents the rate at which the function's values are changing at any given point.

To compute the derivative of the cosine function, one applies the basic differentiation rules. The result is \( f'(x) = -\sin(x) \), which indicates that the derivative of cosine is the negative sine function. This relationship is a part of the trigonometric functions’ derivatives that students must remember for calculus problems. An easy way to remember this is by using mnemonic devices, or by visualizing the unit circle and noticing how the sine function is a 'shift' of the cosine function.

This principle is not only theoretically significant but also practically applicable in various calculus problems, including those involving the Mean Value Theorem.

Applying the Mean Value Theorem

The Mean Value Theorem (MVT) is a critical concept in calculus. It provides a formal way of stating that for any continuous and differentiable function over a certain interval, there's at least one point where the derivative of the function is equal to the average rate of change over that interval.

The formal statement is: if \(f(x)\) is continuous on the closed interval \[a, b\] and differentiable on the open interval \(a, b\), then there exists at least one \(c \ in (a, b)\) such that \(f'(c) = \frac{f(b) - f(a)}{b - a}\).

To apply the MVT, as in the case of \( f(x) = \cos(x) \) for \( x \) between \(0\) and \(\pi / 3\), one must ensure the function meets the conditions of the theorem — it must be continuous and differentiable on the interval in question. In practice, this means calculating the average rate of change over the interval, and then finding the point at which the derivative equals this average — as was done in steps 2 and 3 of the original solution.

Computing Derivatives

Computing derivatives is a core job in calculus. It involves finding the derivative of a function, which is a measure of how a function's output value changes in response to changes in its input value. Essentially, it’s the slope of the function at any point, or the 'instantaneous rate of change'.

When computing derivatives, it's crucial to first identify the type of function you're dealing with, as different rules apply to different functions. For trigonometric functions like the cosine function, specific differentiation formulas apply. As discussed earlier, the derivative of \( \cos(x) \) is \( -\sin(x) \).

Then, to compute a specific value of the derivative, you substitute the given value of \(x\) into the derivative function. For instance, if you want to find the slope of \( \cos(x) \) at \( x = \pi / 6\), you compute \( -\sin(\pi / 6)\). Derivatives are necessary in many fields, from motion in physics to calculating the marginal cost in economics. Mastering the computation of derivatives is thus an essential skill for solving complex problems in calculus and beyond.