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)?

2 Answers
Apr 14, 2018

(a) The tangent is y = -2x+2y=2x+2
(b) The normal is y=x/2+2y=x2+2

Explanation:

a) The slope of the tangent to a curve y=F(x)y=F(x) at (x_0,F(x_0))(x0,F(x0)) is given by F^'(x_0)F(x0). 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)

Apr 14, 2018

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