Question

Find the critical numbers of \(f\) (if any). Find the open intervals on which the function is increasing or decreasing and locate all relative extrema. Use a graphing utility to confirm your results. $$ f(x)=4(x-\arcsin x) $$

Short answer

The given function \(f(x) = 4(x - \arcsin x)\) has one critical number at \(x = 0\), at which point it reaches a local maximum. It is increasing on the interval \((-1, 0)\), and is then constant on the interval \((0, 1)\).

Step-by-step solution

Differentiate the function

Start by finding the derivative of the function. The derivative of \(x\) with respect to \(x\) is 1, and the derivative of \(\arcsin x\) with respect to \(x\) is \( \frac{1}{\sqrt{1 - x^2}} \). Therefore, the derivative \(f'(x)\) can be computed by applying the derivative rules: \(f'(x) = 4(1 - \frac{1}{\sqrt{1 - x^2}})\).

Find the Critical Numbers

Now, to find the critical numbers, we need to find the values of \(x\) where \(f'(x) = 0\) or \(f'(x)\) does not exist. Setting \(f'(x) = 0\), we get \(4(1 - \frac{1}{\sqrt{1 - x^2}}) = 0\) or \(1 = \frac{1}{\sqrt{1 - x^2}}\). Squaring both sides gives \(1 = 1 - x^2\) and solving for \(x\) gives \(x = 0\). This value is within the domain of the function \(\arcsin x\) as \(-1 \leq x \leq 1\), so there is one critical number at \(x = 0\). The derivative of the function is undefined for \(x\) values outside this interval, but these don't need to be considered as critical numbers, as they fall outside the function's domain.

Use a Sign Test to Determine Intervals of Increase / Decrease

To determine where the function is increasing or decreasing, perform a sign test around each critical number. Pick a number less than 0 (e.g., -1) and a number greater than 0 (e.g., 1), and substitute these into \(f'(x)\). If the result is positive, the function is increasing in that interval. If it’s negative, the function is decreasing. For \(x = -1\), \(f'(-1) = 4(1 - \sqrt{2}) > 0\). Therefore, \(f(x)\) is increasing on \((-1, 0)\).For \(x = 1\), \(f'(1) = 4(1 - 1) = 0\), meaning the function is constant; hence it's neither increasing nor decreasing in the interval \((0, 1)\). With this, we can also conclude that the function has a local maximum at \(x = 0\), as the function transitions from increasing to constant at \(x = 0\).

Confirm the Results with Graphing

Lastly, graph the function \(f(x) = 4(x - \arcsin x)\) using a graphing utility and observe that the function increases up to \(x = 0\), where it reaches a maximum, and then remains constant up to \(x = 1\). Thus, the findings match our result.

Key concepts

Derivative of Arcsin

Understanding the derivative of the arcsine function is key for solving calculus problems that involve trigonometric functions. The derivative of \( \arcsin x \) with respect to \( x \) is expressed as \( \frac{1}{\sqrt{1 - x^2}} \). This result comes from the chain rule and the implicit differentiation of the sine function. When you apply the derivative to \( \arcsin x \) in the given function, \( f(x)=4(x-\arcsin x) \), it's crucial to remember domain restrictions for \( \arcsin x \) where \( -1 \leq x \leq 1 \). Calculations outside this range are not valid, which is why they don't impact the determination of critical numbers in the context of this problem.
This knowledge on the derivative of arcsin allows you to adeptly handle step 1 of the solution process, differentiating the composite function with confidence.

Increasing and Decreasing Intervals

Determining where a function increases or decreases is an essential part of analyzing its behavior. By examining the sign of the derivative, \( f'(x) \), you can establish these intervals. If the derivative is positive over an interval, the function is increasing, and if negative, the function is decreasing. In our exercise, after finding the critical number \( x = 0 \), we performed a sign test. This showed that \( f(x) \) is increasing on the interval \( (-1, 0) \), because \( f'(-1) > 0 \). Within the interval \( (0, 1) \) the derivative is zero, so the function doesn’t increase or decrease. This helps us to understand and identify the nature of intervals and predict the behavior of the function just by looking at the sign of its derivative.

Relative Extrema

Relative extrema are the high and low points on a function within a certain interval. They correspond to local maxima or minima. To locate these points, we first find the critical numbers where the derivative is zero or undefined, and then use a sign test or the first derivative test to determine the nature of these points. In our function, \( f(x) = 4(x - \arcsin x) \), we have identified \( x = 0 \) as a critical point. By observing that the function increases before this point and is constant after, we can conclude there is a relative (local) maximum at \( x = 0 \). Understanding relative extrema is crucial for sketching graphs and for optimization problems in calculus.

Sign Test Calculus

A sign test is a simple yet powerful tool in calculus to determine the behavior of functions around critical points. After finding the critical numbers and ensuring they are within the domain of the function, we select values from the intervals around each critical number and substitute them into the derivative. If the derivative value for a chosen test point is positive, the function is increasing at that interval. If it is negative, the function is decreasing. No change in sign indicates a zero derivative or a constant function. In our scenario, the sign test revealed that for \( x < 0 \) the function is increasing and for \( x > 0 \) it remains constant, indicating a local maximum at the critical point \( x = 0 \). This method confirms our understanding of the function’s behavior around the critical points.