Given:
The function
[tex]f(x)=\frac{2x}{x^2+1}[/tex]
Required:
Find the equation of the tangent at f(x) at x=0.
Explanation:
The slope of f(x),
[tex]f^{\prime}(x)=\frac{2(-x^2+1)}{(x^2+1)^2}[/tex]
Now, the slope of function at x = 0 is f'(x) = 2.
The line with slope m = 2 and passing through (0, 0).
[tex]f(x)=2x[/tex]
Answer:
Equation of tangent of the function at x = 0, f(x) = 2x.
