Question

Absolute maxima and minima Determine the location and value of the absolute extreme values of \(f\) on the given interval, if they exist. $$f(x)=\sin 3 x \text { on }[-\pi / 4, \pi / 3]$$

Short answer

Answer: The absolute maximum value is \(1\) at \(x = \frac{\pi}{6}\), and the absolute minimum value is \(-\frac{\sqrt{2}}{2}\) at \(x = -\frac{\pi}{4}\).

Step-by-step solution

Find the derivative of \(f(x)\)

Using the chain rule, we find the derivative of \(f(x)=\sin{3x}\) with respect to \(x\). That is: $$f'(x) = \frac{d}{dx} \sin{3x} = 3 \cos{3x}.$$

Find the critical points

To find the critical points, set \(f'(x)\) equal to zero and solve for \(x\): $$3\cos{3x} = 0 \Rightarrow \cos{3x} = 0.$$ The locations where \(\cos{y}=0\) are at \(y=(2n+1)\frac{\pi}{2}\) for any integer \(n\). In this case, our \(y\) is equal to \(3x\), so $$3x = (2n+1)\frac{\pi}{2} \Rightarrow x = \left(2n+1\right)\frac{\pi}{6}.$$

Determine critical points within the interval

We need to find the critical points within the interval \([-\pi/4, \pi/3]\). Using the formula \(x = \left(2n+1\right)\frac{\pi}{6}\), we check for different values of \(n\). For \(n = -1\), we get \(x = -\frac{\pi}{2}\), which is not in the interval. For \(n = 0\), we get \(x = \frac{\pi}{6}\), which is in the interval. So, x = \(\frac{\pi}{6}\) is a critical point. For \(n=1\), we get \(x = \frac{5\pi}{6}\), which is not in the interval. Also, we must not forget that we need to evaluate the function at the interval endpoints as well, so we include \(x=-\pi/4\) and \(x=\pi/3\).

Evaluate \(f(x)\) at critical points and endpoints

Now let's evaluate \(f(x)\) at the critical points and endpoints: $$f(-\frac{\pi}{4}) = \sin\left(-\frac{3\pi}{4}\right) = -\frac{\sqrt{2}}{2}$$ $$f(\frac{\pi}{6}) = \sin\left(\frac{\pi}{2}\right) = 1$$ $$f(\frac{\pi}{3}) = \sin\left(\pi\right) = 0$$

Conclusion

From the evaluations in Step 4, we can conclude that: The absolute maximum value is \(f(\frac{\pi}{6}) = 1\) at \(x = \frac{\pi}{6}\). The absolute minimum value is \(f(-\frac{\pi}{4}) = -\frac{\sqrt{2}}{2}\) at \(x = -\frac{\pi}{4}\).

Key concepts

Absolute maxima and minima

Absolute maxima and minima refer to the highest and lowest values that a function reaches over a particular interval. These extreme values are essential in calculus as they allow us to evaluate not just specific points but overall behavior.

To find these values, you need to check the endpoints of the interval and any critical points within the interval. Here's a brief process:

• Calculate the derivative of the function to find critical points.
• Set the derivative equal to zero to solve for potential critical points.
• Evaluate the original function at the endpoints and at all critical points found.

Once these values are known, you compare them to determine which one is the absolute maximum and which is the absolute minimum.

Critical points

Critical points are points on the graph of a function where the derivative is zero or undefined. These points are significant because they indicate where a function might have local maxima or minima.

For the function \(f(x) = \sin{3x}\), you'd first find the derivative, which is given as \(f'(x) = 3\cos{3x}\). Solve the equation \(f'(x) = 0\) to find:

• Places where the slope of the tangent is zero, indicating potential maximum or minimum values.

In our example, \(\cos{3x} = 0\) leads us to \(x = \left(2n+1\right)\frac{\pi}{6}\). You then identify critical values within the given interval by testing different integer values for \(n\).

Chain rule

The chain rule is a fundamental technique used in calculus to differentiate composite functions. A composite function is one where a function is applied to the value of another function, such as \(\sin{3x}\) where the inner function is \(3x\) and the outer is \sin{x}\.

To apply the chain rule, follow the process:

• Differentiate the outer function, keeping the inner function unchanged.
• Multiply by the derivative of the inner function.

For example, when differentiating \(\sin{3x}\), we find the derivative of \(\sin{x}\) which is \cos{x}\, and then multiply it by the derivative of \(3x\) which is \3\. This results in \(3\cos{3x}\), demonstrating how the chain rule combines both derivatives.

Trigonometric functions

Trigonometric functions, like \(\sin{x}\) and \(\cos{x}\), are periodic functions that oscillate between set values. They are crucial in many areas of math and science, involving waves, circles, and oscillations.

The sine function, \(\sin{x}\), varies between -1 and 1, and is periodic with a period of \(2\pi\). It is important to understand its behavior over different intervals.

• The \(\sin{x}\) function reaches its maximum value of 1 when \(x = \frac{\pi}{2} + 2k\pi\), where \(k\) is any integer.
• Its minimum value of -1 occurs at \(x = \frac{3\pi}{2} + 2k\pi\).

In the context of the given interval \(\left[-\frac{\pi}{4}, \frac{\pi}{3}\right]\), it's key to understand where these maxima and minima could occur to evaluate absolute extrema effectively.