Question

A) Determine the equation of the tangent line to F(x) = (2 - x^2)/(1+x) at (0,2). write your answer in the form Y=mx+b? b) determine the equation of the normal to f(x) at (0,2)?

Answer 1

(a) The tangent is `y = -2x+2`
(b) The normal is `y=x/2+2`

Explanation:

a) The slope of the tangent to a curve `y=F(x)` at `(x_0,F(x_0))` is given by `F^'(x_0)`. Now

`F^'(x) = ((1+x)d/dx(2-x^2)-(2-x^2)d/dx(1+x))/(1+x)^2`
`qquad = ((1+x)(-2x)-(2-x^2)(1))/(1+x)^2 = -(x^2+2x+2)/(1+x)^2`

So

`F^'(0) = -2`

So, the tangent at `(0,2)` is a straight line passing through `(0,2)` with a slope `-2` and thus is given by

`y-2 = (-2) (x-0)`

or

`color(red)(y=-2x+2)`

b) The normal at `(0,2)` being perpendicular to the tangent there, has a slope of `-1/(-2) = 1/2`, and is hence given by

`y-2 = 1/2(x-0)`

or

`color(red)(y = x/2+2)`

Answer 2

Tangent line y = -2x + 2
Normal line y = `1/2 x + 2`

Explanation:

f(x) = `(2-x^2)/(1+x)` (0,2). `(x_1,y_1)`

f'(x) = `((1+x)(2-x^2)' - (2-x^2)(1+x)')/(1+x)^2`
f'(x) = `((1+x)(-2x) - (2-x^2)(1))/(1+x)^2`
f'(x) = `(-2x - 2x^2 - 2 + x^2) / (1+x)^2`
f'(x) = `(-x^2 -2x - 2)/(1+x)^2`
Log in x = 0 then you get the slope of tangent line m = -2

Tangent line `y-y_1=m(x-x_1)`
`y-2=-2(x-0)` => y= -2x +2

Normal line is the line that perpendicular to the tangent line. Therefore, y-2 = `1/2 (x-0) => y =` `1/2 x+ 2`