Question

Slope of the tangent line?

Find slope of the tangent line to the function:

`f(x)=(3x)/(x+4)`

using `m=lim_(h->0) (f(x+h)-f(x))/h`

Answer 1

The slope of the tangent at the point `(x,{3x}/{x+4})` is `m = 12/(x+4)^2`

Explanation:

`f(x) = {3x}/{x+4} implies`
`f(x+h) = {3(x+h)}/{x+h+4}`

Thus
`f(x+h)-f(x) = {3(x+h)}/{x+h+4}- {3x}/{x+4}`
`qquad = {3(x+h)(x+4)-3x(x+h+4)}/{(x+4)(x+h+4)}`
`qquad = 3 {x^2+(h+4)x+4h-x^2-(h+4)x}/{(x+4)(x+h+4)}={12h}/{(x+4)(x+h+4)}`

So

`{f(x+h)-f(x)}/h = {12}/{(x+4)(x+h+4)}`

and finally

`m = lim_{h to 0} {f(x+h)-f(x)}/h = lim_{h to 0}{12}/{(x+4)(x+h+4)} = 12/(x+4)^2`

Answer 2

Slope is `m=12/(x+4)^2`

Explanation:

Slope of the tangent is given by the first derivative of the function.

Here we have `f(x)=(3x)/(x+4)`

hence slope`m=(df)/(dx)=lim_(h->0)(f(x+h)-f(x))/h`

= `lim_(h->0)((3x+3h)/(x+h+4)-(3x)/(x+4))/h`

= `lim_(h->0)((3x+3h)(x+4)-(3x)(x+h+4))/(h(x+4)(x+h+4)`

= `lim_(h->0)((3x^2+12x+3xh+12h-3x^2-3xh-12x))/(h(x+4)(x+h+4)`

= `lim_(h->0)((12h))/(h(x+4)(x+h+4)`

= `lim_(h->0)12/((x+4)(x+h+4)`

= `12/(x+4)^2`