May be there is an operator missing in the first function, h(x). I will solve this in two ways, 1) as if the h(x) = 5x and 2) as if h(x) = 5 + x
1) If h(x) = 5x and k(x) = 1/x
Then (k o h) (x) = k ( h(x) ) = k(5x) = 1/(5x)
2) If h(x) = 5 + x and k (x) = 1/x
Then (k o h)(x) =k ( h(x) ) = k (5+x) = 1 / [5 + x]
Answer:
infinite solutions
Step-by-step explanation:
Eliminate parentheses and collect terms:
3x -12 +5 -x = 2x -7
2x -7 = 2x -7
This is true for any value of x. There are an infinite number of solutions.
