I know that Dom$ (f) = (-\infty,4]$ Dom $(g) = \mathbb{R}$, but the problem is Dom$(f+g)$.

Please also tell Dom$(f-g)$

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    $\begingroup$ What have you tried? Also please use the latex symbols... Hint : any polynomial function can be defined over $\mathbb{R}$ naturally and square root of a function is well-defined in $\mathbb{R}$ only for nonnegative numbers $\endgroup$ – Evan William Chandra Dec 24 '19 at 7:26

Since $g(x)=x^2$ does exists for every $x in \mathbb R$ and since $\sqrt{4-x}$ require that $4-x \geq 0$, the Domain of $f+g$ is simply

$$ D = \left\{ {x \in \mathbb{R}:x \geqslant 4} \right\} $$

  • $\begingroup$ Thanks, can you please tell me if the domain will remain same for f-g or not. $\endgroup$ – Rajat Dash Dec 24 '19 at 8:21
  • $\begingroup$ Yes, because the "problem" arises from the square root and as long as you add or subtract functions defined for every $x$ in $\mathbb R$ nothing change. $\endgroup$ – Luca Goldoni Ph.D. Dec 24 '19 at 9:17
  • $\begingroup$ Thank you so much. $\endgroup$ – Rajat Dash Dec 24 '19 at 9:29
  • $\begingroup$ You are welcome! $\endgroup$ – Luca Goldoni Ph.D. Dec 24 '19 at 9:35

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