The function f(x) = sin^{-1}(sqrt{x}) is defi | vedclass

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(B) For the function $f(x) = \sin^{-1}(\sqrt{x})$ to be defined,the argument of the inverse sine function must lie in the interval $[-1, 1]$.Thus,we m

Vedclass · Accepted Answer

$[0, 1]$

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(B) For the function $f(x) = \sin^{-1}(\sqrt{x})$ to be defined,the argument of the inverse sine function must lie in the interval $[-1, 1]$.Thus,we must have $-1 \le \sqrt{x} \le 1$.Since $\sqrt{x}$ is always non-negative,the condition $-1 \le \sqrt{x}$ is always satisfied for all $x \ge 0$.Now,we solve $\sqrt{x} \le 1$.Squaring both sides,we get $x \le 1^2$,which implies $x \le 1$.Also,for $\sqrt{x}$ to be defined,we must have $x \ge 0$.Combining these conditions,we get $0 \le x \le 1$.Therefore,the function is defined in the interval $[0, 1]$.

The function $f(x) = \sin^{-1}(\sqrt{x})$ is defined in the interval:

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