Question

Find all relative extrema. Use the Second Derivative Test where applicable. \(f(x)=\arcsin x-2 x\)

Short answer

The relative maximum of the function is at \(\frac{\sqrt{3}}{2}\) and the relative minimum is at \(-\frac{\sqrt{3}}{2}\).

Step-by-step solution

Compute the First Derivative

Find the first derivative of the function \(f(x)\). The derivative of \(\arcsin x\) is \(\frac{1}{\sqrt{1-x^2}}\) and the derivative of \(2x\) is \(2\). Thus, the first derivative \(f'(x)\) is: \(f'(x) = \frac{1}{\sqrt{1-x^2}}-2\).

Find the Critical Points

Set the first derivative equal to zero to find the critical points: \(0 = \frac{1}{\sqrt{1-x^2}}-2\). This gives \(x = \pm \frac{\sqrt{3}}{2}\). Since the absolute value of \(x\) can not be larger than 1, these are valid critical points.

Compute the Second Derivative

Now, compute the second derivative \(f''(x)\). The derivative \(\frac{1}{\sqrt{1-x^2}}\) is \(\frac{x}{(1-x^2)^\frac{3}{2}}\). The derivative of \(2\) is \(0\). So, the second derivative \(f''(x)\) is: \(f''(x) =\frac{x}{(1-x^2)^\frac{3}{2}}\).

Apply the Second Derivative Test

To classify the critical points and find relative extrema, substitute the critical points into \(f''(x)\). If \(f''(x)\) is positive, you have a relative minimum; if \(f''(x)\) is negative, you have a relative maximum. For \(x = \frac{\sqrt{3}}{2}\), \(f''(x)\) is negative, so \(\frac{\sqrt{3}}{2}\) is a relative maximum. For \(x = -\frac{\sqrt{3}}{2}\), \(f''(x)\) is positive, so \(-\frac{\sqrt{3}}{2}\) is a relative minimum.