Question

What is the equation of the tangent line of `f(x) =(1-x^3)/(x^2-3x)` at `x=4`?

Answer 1

`y=(123/16)x-46`

Explanation:

The slope of the tangent line at x=4 is `f'(4)`
let us find `f'(x)`
`f(x)` is in the form `u/v` then
`f'(x)=(u'v-v'u)/v^2`
let `u=1-x^3` and `v=x^2-3x`
So,
`u'=-3x^2`
`v'=2x-3`
then
`f'(x)=(u'v-v'u)/v^2`
`f'(x)=(((-3x^2)(x^2-3x))-((2x-3)(1-x^3)))/(x^2-3x)^2`
`f'(x)=(-3x^4+9x^3-2x+2x^4+3-3x^3)/(x^2-3x)^2`
`f'(x)=(-x^4+6x^3-2x+3)/(x^2-3x)^2`

To find the slope of tangent line at x=4 we need to compute f'(4)
We evaluated f'(x) so lrt us substitute x by 4
`f'(4)=(-4^4+6*4^3-2*4+3)/(4^2-3*4)^2`
`f'(4)=(-256+384-8+3)/(16-12)^2`
`f'(4)=123/16`
The slope of this tangent is 123/16
Having `x=4` let us find `y`
`y=(1-4^3)/(4^2-3*4)`
`y=-63/4`
The equation of the tangent line is :
`y-(-63/4)=123/16(x-4)`
`y+63/4=(123/16)x-123*4/16`
`y+63/4=(123/16)x-123/4`
`y=(123/16)x-123/4-63/4`
`y=(123/16)x-(123+63)/4`
`y=(123/16)x-184/4`
`y=(123/16)x-46`