Question

Absolute maxima and minima Determine the location and value of the absolute extreme values of \(f\) on the given interval, if they exist. $$f(x)=\cos ^{2} x \text { on }[0, \pi]$$

Short answer

The absolute maximum value of the function on the given interval is 1 at \(x=0\) and \(x=\pi\) and the absolute minimum value is 0 at \(x=\frac{\pi}{2}\).

Step-by-step solution

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

To find the critical points of \(f(x)\), we start by first finding the derivative of the given function with respect to \(x\). $$ f(x) = \cos^2 x\\ f'(x) = \frac{d}{dx}(\cos^2 x) $$ Using the chain rule, we get: $$ f'(x) = 2(\cos x)(-\sin x) = -2\sin x\cos x $$

Find the critical points of \(f(x)\)

Critical points are found where the derivative is zero or undefined. In this case, \(f'(x)\) is continuous for all \(x\) and has no undefined points. So, we only need to set it to zero and solve for \(x\). $$ -2\sin x\cos x=0 $$ $$ \sin x\cos x=0 $$ The function \(\sin x\cos x=0\) when either \(\sin x=0\) or \(\cos x=0\). For \(\sin x=0\), we have the points \(x = 0\) and \(x = \pi\). For \(\cos x=0\), we have the points \(x = \frac{\pi}{2}\). These values are our critical points, so we have \(x = 0, \frac{\pi}{2}, \pi.\)

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

Evalute the function \(f(x)\) at the critical points and endpoints of the interval to determine the maximum and minimum values: 1. At \(x=0\), \(f(x)=f(0)=\cos^2{(0)}=1\) 2. At \(x=\frac{\pi}{2}\), \(f(x)=f\left(\frac{\pi}{2}\right)=\cos^2\left(\frac{\pi}{2}\right)=0\) 3. At \(x=\pi\), \(f(x)=f(\pi)=\cos^2{(\pi)}=(\cos{(\pi)})^{2} = (-1)^2 = 1\) For the given interval \([0, \pi]\), \(f(x)\) takes a maximum value of \(1\) at \(x=0\) and \(x=\pi\) and a minimum value of \(0\) at \(x=\frac{\pi}{2}\).

Key concepts

Critical Points

Identifying critical points is an important step when finding absolute maxima and minima of a function. Critical points occur where the derivative of the function is zero or undefined. For continuous functions, like polynomial or trigonometric functions, critical points are often found by setting the derivative to zero. To find critical points: - Compute the derivative of the function, referred to as \(f'(x)\). - Solve the equation \(f'(x) = 0\) to find potential critical points. - Check if any points where the derivative is undefined also occur; for some functions, these could also be critical points. In our example, the derivative \(f'(x) = -2\sin x\cos x\) is set to zero, yielding critical points at \(x = 0, \frac{\pi}{2}, \pi\), since \(\sin x \cos x = 0\) at these values.

Derivative

The derivative of a function provides a lot of insight into its behavior. It tells us how the function is changing at any given point—a positive derivative indicates increasing values, while a negative derivative shows decreasing values.To find the derivative, we often use rules like: - The power rule: For \(x^n\), the derivative is \(nx^{n-1}\).- The product rule: For two functions \(u(x)\) and \(v(x)\), the derivative of their product is \(u'v + uv'\).- The chain rule: For a composite function like \(g(f(x))\), the derivative is \(g'(f(x)) \cdot f'(x)\).In the given exercise, the chain rule is used to differentiate \(f(x) = \cos^2 x\), resulting in \(f'(x) = -2\sin x \cos x\). Here, \(\cos^2 x\) is interpreted as \((\cos x)^2\) which allows us to apply the chain rule.

Interval Notation

Interval notation is the method used to describe a set of numbers lying between two endpoints. It’s a concise way to represent all the numbers between a start and end within a function's domain.Types of interval notations include:

• \([a, b]\): a closed interval, including the endpoints \(a\) and \(b\).
• \((a, b)\): an open interval, excluding the endpoints \(a\) and \(b\).
• \((a, b]\) or \([a, b)\): half-open intervals, including only one endpoint.

The exercise uses the interval notation \([0, \pi]\), indicating a closed interval where both \(0\) and \(\pi\) are included. This is essential when evaluating the function for its absolute extrema, as the behavior at the endpoints can affect the evaluation.