Question

Its all one big question

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

Answer 1

see explanation

Explanation:

h(x) + k(x) = x² + 1 + x - 2 = x² + x - 1

(h + k)(2) = 2² + 2 - 1 = 4 + 2 - 1 = 5

h(x) - k(x) = x² + 1 - (x - 2) = x² + 1 - x + 2 = x² -x + 3

(h - k)(3) = 3² - 3 + 3 = 9 - 3 + 3 = 9

h(2) = 2² + 1 = 4 + 1 = 5

k(3) = 3 - 2 = 1

3h(2) + 2k(3) = (3 × 5) + (2 × 1) = 15 + 2 = 17

Answer 2

(h + k)(2) = 5, (h – k)(3) =9 and 3h(2) + 2k(3) = 17.

Explanation:

The given functions are

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

`k(x)=x-2`

First find the values of functions at x=2 and x=3.

At x=2,

`h(2)=(2)^2+1=5`

`k(2)=2-2=0`

At x=3,

`h(3)=(3)^2+1=10`

`k(3)=3-2=1`

Now, use these values to find the required values.

`(h+k)(2)=h(2)+k(2)\rightarrow 5+0=5`

`(h-k)(3)=h(3)-k(3)\rightarrow 10-1=9`

`3h(2)+2k(3)=3(5)+2(1)\rightarrow 15+2=17`

Therefore, (h + k)(2) = 5, (h – k)(3) =9 and 3h(2) + 2k(3) = 17.