Question

For `f(x) =x/(x^2+1)`, what is the equation of the line tangent to `x =3` and at what point is the tangent line horizontal?

Answer 1

1. Tangent line is: `y=-0.08x+0.54`

2. The tangent line is horizontal at: `x=-1` and `x=1`

Explanation:

To find tangent lines we first have to calculate `f'(x)`

`f(x)=x/(x^2+1)`

`f'(x)=(1*(x^2+1)-2x*x)/(x^2+1)^2`

`f'(x)=(1-x^2)/(x^2+1)^2`

1. To calculate tangent at `x_0=3` we have to calculate `f'(x_0)`

`f'(3)=(1-3^2)/(3^2+1)=-8/10--0.8`

`f(3)=3/(9+1)=3/10=0.3`

Now we have to find `b` for which `y=-0.8x+b` passes through `(3;0.3)`

`0.3=-0.08*3+b`

`b=0.3+0.24`

`b=0.54`

Answer 1: The tangent is `y=-0.8x+0.54`

To solve the second part we have to find points for which `f'(x)=0`

`(1-x^2)/(x^2+1)^2=0`

The fraction is zero if and only if its numerator is zero.

`1-x^2=0`

`x^2=1`

`x=-1 vv x=1`

Answer 2: Tangent lines of `f(x)` are horizontal for `x=-1` and `x=1`