Find (k ∘ g)(x) and (g ∘ k)(x) for k(x)=7x−2 and g(x)=x2−7x+8

Answer 1

Hadassah, I think both Anita have uncharacteristically made typographical errors.

Here’s my turn:

k(g(x)) = 7(g(x)) -2 = 7(x²-7x+8) - 2

= 7x²-49x +54

g(k(x)) = (k(x))² -7k(x) +8

= (7x-2)²-7(7x-2) +8

= 49x²-28x +4 -49x +14 +8

= 49x²- 77x + 26

Answer 2

So, just like k(3) = 7(3) - 2, etc,

k(g(x)) = 7(x² - 7x + 8) - 2 = 7x² - 49x + 56 - 2 = 7x² - 49x + 6

g(k(x)) = (7x - 2)² - 7x + 8 = 49x² - 28x + 4 - 7x + 8 = 49x² - 35x + 12

Answer 3

Hello Hadassah,

When working with composition of functions, you may find a starter step with something like this example.

Let f(x) = 3x - 2

Then f(2t - 1) looks like this: f(2t - 1) = 3(2t - 1) - 2

= 6t - 3 - 2

f(2t - 1) = 6t - 5

In your problem, (kοg)x = 7(g(x)) - 2 We are inserting g(x) for the x in k(x).

= 7(x² - 7x + 8)

(kοg)x = 7x² - 49x +56

The (gοk)x starts in a similar manner by "embedding" k(x) into g(x). This will be a little more involved algebraically, but the strategy is the same.

(gοk)x = k² - 7k + 8 We are inserting k(x) for the x in g(x).

= (7x - 2)² - 7(7x - 2) + 8

= 49x² - 28x + 4 + 8

(gοk)x = 49x² - 28x + 12

Regards,

Anita A.