How do you find the domain of $f \circ g$ $f @ g$ given $f (x) = \sqrt{x - 1}$ $f(x)= sqrt(x-1)$ and $g (x) = \frac{1}{x - 7}$ $g(x)= 1/(x-7)$ ?

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Answer 1 · 2018-06-14T19:02:50

$x \in (7, 8]$ $x in (7, 8]$

$f (\frac{1}{x - 7}) = \sqrt{\frac{1}{x - 7} - 1}$ $f(1/(x-7)) = sqrt{1/(x-7) - 1}$

$x \neq 7 and \frac{1}{x - 7} - 1 \geq 0$ $x ne 7 and 1/(x-7) - 1 >= 0$

$\frac{1 - x + 7}{x - 7} \geq 0$ $frac{1 - x + 7}{x - 7} >= 0$

$(x - 7) (x - 8) \leq 0$ $(x - 7)(x - 8) <= 0$

$7 \leq x \leq 8$ $7 <= x <= 8$

How do you find the domain of f @ gf∘gf @ g given f(x)= sqrt(x-1)f(x)=√x−1f(x)= sqrt(x-1) and g(x)= 1/(x-7)g(x)=1x−7g(x)= 1/(x-7)?