Question

the function $f(x)=\frac{8}{x - arcsin(x)}sqrt{63}$ has two critical values. determine these two critical values. enter the smaller critical value in the first blank and the larger critical value in the second blank. smaller critical value: larger critical value:

Explanation:

Step1: Recall the domain of arcsin(x)

The domain of \(y = \arcsin(x)\) is \(- 1\leqslant x\leqslant1\). Also, for the function \(f(x)=\frac{8}{x}-\arcsin(x)\), we need to consider the denominator \(x
eq0\). To find critical - points, we first find the derivative of \(f(x)\). The derivative of \(\frac{8}{x}=8x^{-1}\), and its derivative using the power rule \((x^n)^\prime=nx^{n - 1}\) is \(-\frac{8}{x^{2}}\), and the derivative of \(\arcsin(x)\) is \(\frac{1}{\sqrt{1 - x^{2}}}\). So \(f^\prime(x)=-\frac{8}{x^{2}}-\frac{1}{\sqrt{1 - x^{2}}}\). Critical points occur where \(f^\prime(x) = 0\) or \(f^\prime(x)\) is undefined. \(f^\prime(x)\) is undefined when \(x = 0\) (because of the \(\frac{8}{x^{2}}\) term) and when \(x=\pm1\) (because of the \(\frac{1}{\sqrt{1 - x^{2}}}\) term). But \(x = 0\) is not in the domain of \(f(x)\).

Step2: Evaluate the behavior at the domain - boundary points

We consider the domain of \(f(x)\) which is \(-1\leqslant x\lt0\) or \(0\lt x\leqslant1\). We check the values of \(x=-1\) and \(x = 1\).

Answer:

Smaller Critical Value: \(-1\)
Larger Critical Value: \(1\)