I'm assuming that you meant:
1
f(x) = -------- and that you want to find the value of x at which f(x) = h(x).
x+1
Of course you could create a table for each f(x) and h(x), but setting f(x)=h(x) and solving for x algebraically would be faster and more efficient:
1
f(x) = -------- = 2x + 3 = h(x). Then 1 = (x+1)(2x+3) = 2x^2 + 3x + 2x + 3
x+1
or 1 = 2x^2 + 5x + 3, or 2x^2 + 5x + 2 = 0.
This is a quadratic equation with a=2, b=5 and c=2. The discriminant is b^2-4ac, or 5^2-4(2)(2), OR 25-16= 9.
Thus, the roots are
-5 plus or minus sqrt(9)
x = ------------------------------------
2(2)
-5 plus or minus 3
= ----------------------------------
4
= {-1/2, -2}
Thus, f(x) = h(x) at both x=-1/2 and x= -2.
