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

1 Answer

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

desmos.comdesmos.com

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