Question

H(x) = x2 + 1 k(x) = x – 2 (h + k)(2) = (h – k)(3) = Evaluate 3h(2) + 2k(3) =

Answer 1

For this case we have the following functions:

`h (x) = x ^ 2 + 1`

`k (x) = x - 2`

We must perform each of the operations shown.

We have then:

For (h + k) (2):

`(h + k) (x) = h (x) + k (x)`

`(h + k) (x) = (x ^ 2 + 1) + (x - 2)`

`(h + k) (x) = x ^ 2 + x - 1`

Evaluating for x = 2 we have:

`(h + k) (2) = 2 ^ 2 + 2 - 1`

`(h + k) (2) = 4 + 2 - 1`

`(h + k) (2) = 5`

For (h - k) (3):

`(h - k) (x) = h (x) -k (x)`

`(h - k) (x) = (x ^ 2 + 1) - (x - 2)`

`(h - k) (x) = x ^ 2 - x + 3`

Evaluating for x = 3 we have:

`(h - k) (3) = 3 ^ 2 - 3 + 3`

`(h - k) (3) = 9 - 3 + 3`

`(h - k) (3) = 9`

For 3h (2) + 2k (3):

`3h (2) + 2k (3) = 3 (2 ^ 2 + 1) + 2 (3 - 2)`

`3h (2) + 2k (3) = 3 (5) + 2 (1)`

`3h (2) + 2k (3) = 15 + 2`

`3h (2) + 2k (3) = 17`

Answer:

`(h + k) (2) = 5`

`(h - k) (3) = 9`

`3h (2) + 2k (3) = 17`