Question

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

Answer 1

`y=(1/2-ln 4)x+1/2-ln 16`

Explanation:

the given function
`f(x)=(ln (x^2+3))/x`
at `x=-1`

Solve for the point first

`f(-1)=(ln ((-1)^2+3))/(-1)`

`f(-1)=y=-ln 4`

the point of tangency `(-1, -ln 4)`

Solve for the slope `m=f' (-1)`

take the first derivative

`f' (x)=(x*d/dx(ln(x^2+3))-ln(x^2+3)*d/dx(x))/(x^2)`

`f' (x)=(x*1/(x^2+3)*(2x+0)-ln(x^2+3)(1))/(x^2)`

Compute slope `f' (-1)`

`f' (-1)=((-1)(1/((-1)^2+3))(2(-1)+0)-ln((-1)^2+3)(1))/((-1)^2)`

`f' (-1)=1/2-ln 4`

Write the equation of the line using Point-Slope Form

`y-(-ln 4)=(1/2-ln 4)*(x-(-1))`

`y=(1/2-ln 4)*x+1/2-ln 16`

Kindly see the graph for better explanation

God bless....I hope the explanation is useful.