Answer:
Step-by-step explanation:
1.a.
[tex]f(x)=\dfrac{5}{x} \ so\\\dfrac{f(x)-f(a)}{x-a}=\dfrac{\dfrac{5}{x}-\dfrac{5}{a}}{x-a}=\dfrac{\dfrac{5*a-5*x}{x*a}}{x-a}\\ = \dfrac{\dfrac{-5(x-a)}{x*a}}{x-a}=\dfrac{-5}{ax}[/tex]
b.
[tex]\dfrac{f(x+h)-f(x)}{h}=\dfrac{\dfrac{5}{x+h}-\dfrac{5}{x}}{h}=\dfrac{\dfrac{5(x-x-h)}{(x+h)x}}{h}=\dfrac{-5h}{(x+h)xh}=\dfrac{-5}{(x+h)x}[/tex]
2.a.
[tex]f(x)=\dfrac{1}{x-1} \ for \ x \ different \ from \ 1 \ \ \ so\\\dfrac{f(x)-f(a)}{x-a}=\dfrac{\dfrac{1}{x-1}-\dfrac{1}{a-1}}{x-a}=\dfrac{\dfrac{a-1-x+1}{(x-1)(a-1)}}{x-a}\\ = \dfrac{\dfrac{-(x-a)}{(x-1)(a-1)}}{x-a}=\dfrac{-1}{(x-1)(a-1)}[/tex]
b.
[tex]\dfrac{f(x+h)-f(x)}{h}=\dfrac{\dfrac{1}{x+h-1}-\dfrac{1}{x-1}}{h}=\dfrac{\dfrac{x-1-x-h+1}{(x+h-1)(x-1)}}{h}=\dfrac{-h}{(x+h-1)(x-1)h}=\dfrac{-1}{(x+h-1)(x-1)}[/tex]
3.a.
[tex]f(x)=\dfrac{1}{x^2} \ so\\\dfrac{f(x)-f(a)}{x-a}=\dfrac{\dfrac{1}{x^2}-\dfrac{1}{a^2}}{x-a}=\dfrac{\dfrac{a^2-x^2}{x^2a^2}}{x-a}\\ = \dfrac{\dfrac{(a-x)(a+x)}{x^2a^2}}{x-a}=\dfrac{-(a+x)}{a^2x^2}[/tex]
b.
[tex]\dfrac{f(x+h)-f(x)}{h}=\dfrac{\dfrac{1}{(x+h)^2}-\dfrac{1}{x^2}}{h}=\dfrac{\dfrac{x^2-(x+h)^2}{(x+h)^2x^2}}{h}=\dfrac{(x-x-h)(x+x+h)}{(x+h)^2x^2h}=\dfrac{-(2x+h)}{(x+h)^2x^2}[/tex]
hope this helps
