The slope of the tangent line to a specific curve in some given point is given by the derivative of that function evaluated in the x-value of that point.
The equation of the tangent line is:
[tex]y = 0.25*x + 0.5[/tex]
Here the function is:
[tex]y = \frac{2x}{x + 2}[/tex]
And the point is (2, 1)
To get the slope we can use the chain rule for derivation:
[tex]f(x) = g(x)*h(x)\\\\f'(x) = g'(x)*h(x) + g(x)*h'(x)[/tex]
Then we can write our function as:
[tex]f(x) = \frac{2x}{x + 2} = g(x)*h(x)\\\\f'(x) = \frac{2}{x + 2} - 2x*\frac{1}{(x + 2)^2}[/tex]
To get the slope at (2, 1) we need to evaluate the function f'(x) at x = 2, we will get:
[tex]f'(2) = \frac{2}{2 + 2} - 2*2*\frac{1}{(2 + 2)^2} = 0.25[/tex]
Then the equation of the tangent line to that point will be:
[tex]y = 0.25*x + b[/tex]
To find the value of b we can use the fact that this line passes through the point (2, 1), then:
[tex]1 = 0.25*2 + b\\1 = 0.5 + b\\1 - 0.5 = b\\0.5 = b[/tex]
Finally, the equation of the tangent line is:
[tex]y = 0.25*x + 0.5[/tex]
If you want to learn more you can read:
https://brainly.com/question/23265136
