Question

a. Find the critical points of \(f\) on the given interval. b. Determine the absolute extreme values of \(f\) on the given interval when they exist. c. Use a graphing utility to confirm your conclusions. $$f(x)=x^{2}+\cos ^{-1} x \text { on }[-1,1]$$

Short answer

Question: Determine the absolute minimum and maximum values of the function \(f(x)=x^2+\cos^{-1}x\) on the interval \([-1,1]\). Answer: The absolute minimum value of the function is \(1+\pi\) at \(x=-1\), and the absolute maximum value is \(1\) at \(x=1\).

Step-by-step solution

Find the critical points

First, we need to find the derivative of the function \(f(x)=x^2+\cos^{-1}x\). The derivative of \(x^2\) is \(2x\), and the derivative of \(\cos^{-1}x\) is \(-1/\sqrt{1-x^2}\) (using the chain rule). So, the derivative of the function is: $$f'(x) = 2x - \frac{1}{\sqrt{1-x^2}}$$ To find critical points, we need to find the values of x where the derivative is either 0 or undefined. We will first solve for \(f'(x)=0\). However, due to the complexity of the equation, it might be hard to solve it analytically. Thus, it's better to investigate the derivative's behavior using the interval of x values.

Determine absolute extreme values

We will now investigate the behavior of the derivative \(f'(x)\) and the function \(f(x)\) at the end points and points in the open interval of \([-1,1]\). Specifically, we want to determine if an absolute minimum or maximum exists. For the two end points, we have: \(f(-1) = -1^2 + \cos^{-1}(-1) = 1 + \pi\) \(f(1) = 1^2 + \cos^{-1}(1) = 1\) As \(f'(x) > 0\) on the open interval (since derivative is defined and positive for \(x\in(-1,1)\)), we can conclude that \(f(x)\) is increasing on \((-1,1)\). At x=-1, the function changes from decreasing (\(f'(-1)\) is undefined) to increasing (\(f'(x)>0\) for \(x\in(-1,1)\)). Therefore, we can conclude that there is an absolute minimum at \(x=-1\). On the other hand, at x=1, the function changes from increasing (\(f'(x)>0\) for \(x\in(-1,1)\)) to undefined (\(f'(1)\) is undefined). Therefore, we can conclude that there is an absolute maximum at \(x=1\). So, the absolute minimum value is \(f(-1)=1+\pi\), and the absolute maximum value is \(f(1)=1\).

Confirm conclusions using a graphing utility

Finally, using a graphing utility (such as Desmos or a graphing calculator), plot the function \(f(x)=x^2+\cos^{-1}x\) on the interval \([-1,1]\). The graph should show a minimum point at \(x=-1\) with the value \(1+\pi\) and a maximum point at \(x=1\) with the value \(1\). This confirms our conclusions.

Key concepts

Extreme Values

When analyzing functions, we often want to find their extreme values. Extreme values consist of the highest and lowest points a function can reach within a given interval. These include absolute or global maximum and minimum values. To find these values, we typically consider both the end points of the interval and any critical points within the interval.

For our function, \(f(x) = x^2 + \cos^{-1}x\), on the interval \([-1, 1]\), we start by evaluating the function at the boundaries of the interval. Here, \(f(-1) = 1 + \pi\) and \(f(1) = 1\). These values offer potential extrema.

It's also important to check any critical points within the interval. As suggested by the derivative's behavior, there are no other points within \((-1, 1)\) where extreme values occur due to \(f'(x) > 0\), indicating the function is continuously increasing. Thus, the absolute minimum is at \(x = -1\) with the value \(1 + \pi\), and the absolute maximum is at \(x = 1\) with the value \(1\).

Derivative Analysis

In calculus, derivative analysis helps us understand the behavior of a function. By analyzing derivatives, we can identify critical points and determine where a function is increasing or decreasing.

For the function \(f(x) = x^2 + \cos^{-1}x\), the derivative \(f'(x) = 2x - \frac{1}{\sqrt{1-x^2}}\) was computed. Critical points occur where the derivative is zero or undefined. In this case, when analyzing \(f'(x)\), we see that it is undefined at the endpoints, \(x=-1\) and \(x=1\), but the function is defined there.
The interval \((-1, 1)\) is crucial since the derivative is positive, indicating that \(f(x)\) is increasing. At these endpoints, the derivative shifts from undefined to defined with a consistent trend across the interval. This behavior suggests that the function has no other critical points inside the interval, allowing us to confirm that extremal values occur at the boundary points.

Interval Notation

Interval notation is a handy way to describe sets of numbers between two endpoints, often used in mathematics to specify domain or range of a function.

For our function \(f(x) = x^2 + \cos^{-1}x\), the interval is given as \([-1, 1]\). The square brackets \([\,]\) indicate that both end points \(-1\) and \(1\) are included. This is crucial, as these endpoints can significantly contribute to finding absolute extremes.
When using interval notation, it helps to swiftly display the range in which we search for critical points and extreme values. Because our function is increasing on \((-1, 1)\), the absolute minimum and maximum values are found right at these closed interval endpoints. This makes interval notation invaluable for quickly identifying these areas without lengthy explanation.