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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?

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6 Answers 6

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

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

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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 $$y\in[0,\sqrt 7]$$

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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.

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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 $[0,\sqrt{7}]$.

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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.

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