This is just the plugging in part...
\(g(x) = 5-\sqrt{4-x}\)
To find g(3 + h), replace every instance of x with (3 + h) .
\(g(3+h) = 5-\sqrt{4-(3+h)} \\g(3+h) =5-\sqrt{4-3-h} \\g(3+h) = 5-\sqrt{1-h}\)
To find g(3), replace every insance of x with 3.
\(g(3) = 5-\sqrt{4-3} \\g(3) =5-\sqrt{1} \\ g(3) =4\)
So
\(m_{tan}=\lim_{h\rightarrow 0} \frac{g(3+h)-g(3)}{h} =\lim_{h\rightarrow 0} \frac{(5-\sqrt{1-h})-(4)}{h} \\~\\=\lim_{h\rightarrow 0} \frac{1-\sqrt{1-h}}{h}\)
Then you gotta actually evaluate the limit and all that to get the slope of the line when x = 3
