Calculus of a Single Variable, Chapter 2, 2.2, Section 2.2, Problem 65

The original equation will intersect a point on the tangent line.
Set both equations equal to each other.
k/x = -3/4 x +3
We have 2 unknowns and 1 equation.
Take the derivative of f(x) in order to find another relationship between k and x.
f(x)=k/x
f(x)=kx^(-1)
f'(x) = -kx^-2
f'(x)=-k/x^2
Substitute the slope -3/4 into f'(x) .
-3/4=-k/x^2
Cross multiply.
3x^2=4k
k=(3x^2) /4
Substitute k back to the original equation to solve for x.
k/x = -3/4 x +3
k* 1/x = -3/4 x +3
((3x^2)) /4 * 1/x = -3/4 x +3
3/4 x= -3/4 x+3
6/4 x = 3
3/2 x = 3
3x = 6
x=2
Substitute the x value back to .
k=(3(2)^2) /4 = (3*4) /4
The answer is: k=3