Question

How do you find `f(a), f(a+h)`, and the difference quotient `(f(a+h) - f(a))/h` of `f(x) = 5/(x+7)` where `h!=0`?

Answer 1

`f(a)=5/(a+7)color(white)("xxx")f(a+h)=5/(a+h+7)color(white)("xxx")(f(a+h)-f(a))/h=-5/((a+7)^2+(a+7)h)`

Explanation:

To find `f(a)` and `f(a+h)`
replace `x` in `f(x)=5/(x+7)` with `a` and `a+h` respectively.

`f(a+h)-f(a)=5/(a+h+7)-5/(a+7)`

`color(white)("XXX")=(5(a+7)-5(a+h+7))/((a+h+7)(a+7))`

`color(white)("XXX")=(-5h)/(a^2+ah+7a+7a+7h+49)`

`color(white)("XXX")=(-5h)/(a^2+14a+49+ah+7h)`

`color(white)("XXX")=(-5h)/((a+7)^2+(a+7)h)`

`(f(a+h)-f(a))/h=((-5h)/((a+7)^2+(a+7)h))/h`

`color(white)("XXX")=(-5)/((a+7)^2+(a+7)h)`

Normally at this point we would be taking the `lim_(hrarr0)` so the `(a+7)h` term in the denominator would effectively disappear.