Problem 16 - Calculus Single Variable

Question

Find a point \(x\) where \(f^{\prime}(x)=6\). $$ f(x)=12 \sin (x) $$

Step-by-Step Solution

Answer

The points are at \(x = \frac{\pi}{3} + 2n\pi\) and \(x = -\frac{\pi}{3} + 2m\pi\), where \(n\) and \(m\) are integers.

Step 1: Find the derivative of the function

To find a point where the derivative equals 6, we first need to find the derivative of the function. For the function \(f(x) = 12 \sin(x)\), the derivative is computed using the derivative of \(\sin(x)\), which is \(\cos(x)\). Hence, the derivative is \(f'(x) = 12 \cos(x)\).

Step 2: Set the derivative equal to 6

We need to find \(x\) such that \(f'(x) = 6\). Substituting the derivative we found, this gives us the equation \(12 \cos(x) = 6\).

Step 3: Solve for \(\cos(x)\)

Divide both sides of the equation \(12 \cos(x) = 6\) by 12 to solve for \(\cos(x)\). This simplifies to \(\cos(x) = \frac{1}{2}\).

Step 4: Determine the angle \(x\)

Now, solve for \(x\) by determining the angles where \(\cos(x) = \frac{1}{2}\). The general solutions are \(x = \frac{\pi}{3} + 2n\pi\) and \(x = -\frac{\pi}{3} + 2m\pi\), where \(n\) and \(m\) are integers. These are because \(\cos(x)\) is \(\frac{1}{2}\) at \(60^{\circ}\) (or \(\frac{\pi}{3}\)) and \(300^{\circ}\) (or \(-\frac{\pi}{3}\)).

Key Concepts

Derivative

In calculus, the derivative of a function essentially measures how the function changes as its input changes. For the function given in our exercise, the derivative signifies how the value of the function changes with a little change in the value of \(x\). It's like checking the speed at which your function is climbing or descending. To derive a function, we follow certain rules depending upon the type of function.The function in this problem is \(f(x) = 12 \sin(x)\). The derivative of \(\sin(x)\) is \(\cos(x)\). Thus, when we take the derivative of \(f(x)\), we multiply the derivative of \(\sin(x)\) by 12, leading to \(f'(x) = 12 \cos(x)\). This tells us that the rate of change of the function is governed not just by the cosine value but also gets amplified 12 times. Derivatives are powerful as they help us determine stationary points and analyze the behavior of graphs.

Derivatives provide insights into how graph function slopes behave. They denote the tangent slope at a point. Key derivations:

• The derivative of \(\sin(x)\) is \(\cos(x)\).
• The derivative of \(12 \sin(x)\) becomes \(12 \times \cos(x)\).
• Derivatives help in finding where the function increases or decreases.

Trigonometric Functions

Trigonometric functions like sine and cosine are essential in calculus, and in understanding periodic phenomena. These functions, based on the angles of a right triangle, are used extensively across sciences to model wave-like behaviors in everything from light and sound to the fluctuating prices of stocks.

The function in our exercise is \(\sin(x)\), multiplied by a constant. When taking the derivative, we rely on the basic trig function derivatives:

• \(\sin(x)\) becomes \(\cos(x)\).
• \(\cos(x)\) becomes \(-\sin(x)\).

In the problem, our derivative comes out to \(12 \cos(x)\), where the number 12 acts as a scalar that scales up the amplitude of the derivative.

Trigonometric properties are especially valuable:

• They help determine angles and distances.
• They are foundational in constructing periodic function graphs.
• Key angle values, like \(\frac{\pi}{3}\) or \(60^{\circ}\), are crucial in solving such problems.

Solving Equations

Solving equations often involves finding the value of a variable that makes a mathematical statement true. In calculus, once we've found our derivative, the next challenge is to identify points at which certain conditions are satisfied. Essentially, this is about setting the expression involving \(x\) to a desired value, and then solving for \(x\).

In the exercise, we need \(f'(x) = 6\). With an equation \(12 \cos(x) = 6\), our goal is to manipulate it to find \(x\).To make \(\cos(x) = \frac{1}{2}\), divide both sides by 12. This simplifies our task into finding angles corresponding to \(\cos(x) = \frac{1}{2}\). Angles where this holds true in a unit circle are specific and repeat every full rotation. The solutions include:

• \(x = \frac{\pi}{3} + 2n\pi\)
• \(x = -\frac{\pi}{3} + 2m\pi\)

where \(n\) and \(m\) are integers representing cycles on the unit circle. Each solution represents a place where the derivative achieves the value 6, and logs periodic repetitions in consistent intervals.