# Find the range of the function $y = \sqrt{7 - x^2}$?

I tried to find the range of the following function : $$y=\sqrt{7-x^2}$$ I found the domain which is: $$y \in (-\sqrt7,\sqrt7)$$ and then I tried to find the range of the function with the method of finding the domain of its inverse function. So : $$y=\sqrt{7-x^2} \Rightarrow$$ $$y^2=7-x^2\Rightarrow$$ $$x^2=7-y^2\Rightarrow$$ $$x=\sqrt{7-y^2}$$ Domain of which is $$x \in (-\sqrt7,\sqrt7)$$ But the correct solution is: $$x \in (0,\sqrt7)$$ Can someone explain where is the mistake?

Since domain of $$y$$ is $$x \in [-\sqrt7 , \sqrt7]$$, then range of $$y$$ is $$y \in [0,\sqrt7]$$.

Since $$\sqrt{7-x^2}$$ can never be negative this excludes all the values $$(-\sqrt{7},0)$$.

I think you have displaced the concept of domain and range. The domain actually constrains the independent variable which is $$x$$ here. So the domain is $$x\in[-\sqrt 7,\sqrt 7]$$and the range consecutively becomes $$y\in[0,\sqrt 7]$$

First, a tidbit about notation. The domain variable is $$x,$$ not $$y,$$ and vice versa. As for the range, recall that if $$y=\sqrt{7-x^2},$$ then since RHS is never negative, it follows that $$0\le y.$$ Combine this with the your calculation that $$|y|\le\sqrt 7,$$ and you're done.

The image of the function described by $$y = \sqrt{7 - x^{2}}$$ is the upper semicircle centered at the origin with radius $$\sqrt{7}$$. Thus its range is given by the set $$[0,\sqrt{7}]$$.

The implicit domain of the function $$y = \sqrt{7 - x^2}$$ is the set of all real numbers $$x$$ such that $$7 - x^2 \geq 0$$. \begin{align*} 7 - x^2 & \geq 0\\ 7 & \geq x^2\\ \sqrt{7} & \geq |x| \end{align*} Hence, $$-\sqrt{7} \leq x \leq 7$$, so the domain of the function is $$[-\sqrt{7}, \sqrt{7}]$$. Notice that the endpoints are included.

The notation $$y = \sqrt{x}$$ denotes the principal (nonnegative) square root of $$x$$.

Therefore, $$y = \sqrt{7 - x^2} \geq 0$$ for each $$x$$ in its domain. In particular, equality holds if $$x = \pm \sqrt{7}$$. Moreover, $$y = \sqrt{7 - x^2}$$ is at most $$\sqrt{7}$$, which occurs when $$x = 0$$. Since $$y = \sqrt{7 - x^2}$$ is continuous on its domain, $$y$$ assumes every value in the closed interval $$[0, \sqrt{7}]$$ by the Intermediate Value Theorem. Hence, the range of the function is $$[0, \sqrt{7}]$$. Again, notice that the endpoints are included.