If f(x)=ax+b and g(x) = cx + d then h(x)=(ax+b)(cx+d).

If f(x) and g(x) are to be tangent to h(x) then they intersect h(x) at a common point.

Therefore I will need to find their points of intersection with h(x):

h(x) and f(x): h(x) and g(x):

(ax+b)(cx+d)=(ax+b) (ax+b)(cx+d)=(cx+d)

(ax+b)(cx+d) – (ax+b) = 0 (ax+b)(cx+d) – (cx+d) = 0

(ax+b)(cx+d – 1) = 0 (cx+d)(ax+b – 1) = 0

ax+b = 0 and (cx+d – 1) = 0 cx+d = 0 and (ax+b – 1) = 0

x = -b/a and x = (1 – d)/c x = -d/c and x = (1 – b)/a

These answers give two x-values for the points of tangency of each function to h(x).

Therefore I need to set them equal to each other to find the one x-value that is a point of tangency for each function to h(x).

-b/a = (1 – d)/c -d/c = (1 – b)/a

-cb = a – ad -ad = c – cb

a = ad – cb c = -ad + cb

Therefore a = -c

This verifies the calculus result that the slopes are opposite of each other.

We can substitute to verify that the y-intercepts have a sum of 1.