Find k such that (-3/4)x + 3 is tangent to the graph of k/x

Question

I have been unable to figure out how to find k

Answer 1

The slope of y = (-3/4)x + 3 is -3/4.

 

If f(x) = k/x, we need to find the value of k so that f'(x) = -3/4.  So, -k/x² = -3/4

 

                                                                                                      k = 3x²/4

 

The curves intersect if (-3/4)x + 3 = k/x

 

                                 (-3/4)x + 3 = (3x²/4)/x = (3/4)x

 

                                             So, (3/2)x = 3

 

                                                          x = 2

 

k = 3x²/4 = 3

 

At the point (2, 3/2), the line y = (-3/4)x + 3 is tangent to the graph of y = 3/x.