Given
The demand function

To determine:
a) The revenue function.
b) The marginal revenue.
c) The marginal revenue when x=200.
d) The equation of tangent, and its derivation.
Explanation:
It is given that,

a) The revenue function is given by,

b) The marginal revenue function is,

c) The marginal revenue when x=200 is,

Hence, the marginal revenue is 3.84.
d) Let y=mx+c is the tangent.
Then,

That implies, for y=12.68, and x=200,

Hence, the equation of tangent is,