Find g’(9) where g(x)=(squareroot(x)+3)/(squareroot(x)-1) ?

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Answer 1 · 2018-02-16T23:08:06

3

$g (x) = \frac{\sqrt{x} + 3}{\sqrt{x} - 1}$ $g(x)=(sqrtx+3)/(sqrtx -1)$

Now substitute $9$ $color(red)(9)$ for every time you see $x$ $color(red)x$ in the expression:

$g (9) = \frac{\sqrt{9} + 3}{\sqrt{9} - 1}$ $g(color(red)9) = (sqrtcolor(red)9+3)/(sqrtcolor(red)9 -1)$

and simplify:

$g (9) = \frac{3 + 3}{3 - 1}$ $g(9)=(3+3)/(3-1)$

$g (9) = \frac{6}{2}$ $g(9)=6/2$

$g (9) = 3$ $g(9)=3$