Question

Help me find this asap

Answer 1

Answer:

DQ = $\frac{5}{(2-\sqrt{x})(2-\sqrt{x+h})(\sqrt{x}-\sqrt{x+h})}$

Step-by-step explanation:

Given function is f(x) = $\frac{5}{2-\sqrt{x}}$

f(x + h) = $\frac{5}{2-\sqrt{x+h} }$

Therefore, indicated difference quotient will be,

DQ = $\frac{f(x+h)-f(x)}{h}$

Now we substitute the values in the difference quotient,

DQ = $\frac{\frac{5}{2-\sqrt{x+h} }-\frac{5}{2-\sqrt{x}}}{h}$

      = $\frac{5(2-\sqrt{x})-5(2-\sqrt{x+h})}{h(2-\sqrt{x})(2-\sqrt{x+h})}$

      = $\frac{10-5\sqrt{x}-10+5\sqrt{x+h}}{h(2-\sqrt{x})(2-\sqrt{x+h})}$

      = $\frac{-5\sqrt{x}+5\sqrt{x+h}}{h(2-\sqrt{x})(2-\sqrt{x+h})}$

      = $\frac{5(-\sqrt{x}+\sqrt{x+h})}{h(2-\sqrt{x})(2-\sqrt{x+h})}$

      = $\frac{5(-\sqrt{x}+\sqrt{x+h})(\sqrt{x}+\sqrt{x+h})}{h(2-\sqrt{x})(2-\sqrt{x+h})(\sqrt{x}+\sqrt{x+h})}$

      = $\frac{5[(x+h)-x]}{h(2-\sqrt{x})(2-\sqrt{x+h})(\sqrt{x}+\sqrt{x+h})}$

      = $\frac{5h}{h(2-\sqrt{x})(2-\sqrt{x+h})(\sqrt{x}+\sqrt{x+h})}$

      = $\frac{5}{(2-\sqrt{x})(2-\sqrt{x+h})(\sqrt{x}+\sqrt{x+h})}$