Answer:
DQ = [tex]\frac{5}{(2-\sqrt{x})(2-\sqrt{x+h})(\sqrt{x}-\sqrt{x+h})}[/tex]
Step-by-step explanation:
Given function is f(x) = [tex]\frac{5}{2-\sqrt{x}}[/tex]
f(x + h) = [tex]\frac{5}{2-\sqrt{x+h} }[/tex]
Therefore, indicated difference quotient will be,
DQ = [tex]\frac{f(x+h)-f(x)}{h}[/tex]
Now we substitute the values in the difference quotient,
DQ = [tex]\frac{\frac{5}{2-\sqrt{x+h} }-\frac{5}{2-\sqrt{x}}}{h}[/tex]
= [tex]\frac{5(2-\sqrt{x})-5(2-\sqrt{x+h})}{h(2-\sqrt{x})(2-\sqrt{x+h})}[/tex]
= [tex]\frac{10-5\sqrt{x}-10+5\sqrt{x+h}}{h(2-\sqrt{x})(2-\sqrt{x+h})}[/tex]
= [tex]\frac{-5\sqrt{x}+5\sqrt{x+h}}{h(2-\sqrt{x})(2-\sqrt{x+h})}[/tex]
= [tex]\frac{5(-\sqrt{x}+\sqrt{x+h})}{h(2-\sqrt{x})(2-\sqrt{x+h})}[/tex]
= [tex]\frac{5(-\sqrt{x}+\sqrt{x+h})(\sqrt{x}+\sqrt{x+h})}{h(2-\sqrt{x})(2-\sqrt{x+h})(\sqrt{x}+\sqrt{x+h})}[/tex]
= [tex]\frac{5[(x+h)-x]}{h(2-\sqrt{x})(2-\sqrt{x+h})(\sqrt{x}+\sqrt{x+h})}[/tex]
= [tex]\frac{5h}{h(2-\sqrt{x})(2-\sqrt{x+h})(\sqrt{x}+\sqrt{x+h})}[/tex]
= [tex]\frac{5}{(2-\sqrt{x})(2-\sqrt{x+h})(\sqrt{x}+\sqrt{x+h})}[/tex]
