```latex
[tex]$
\begin{array}{l}
h(x) = x^2 + 1 \\
k(x) = x - 2 \\
(h + k)(2) = \square
\end{array}
$[/tex]

[tex]$(h - k)(3) = \square$[/tex]

Evaluate [tex]$3h(2) + 2k(3) = \square$[/tex]
```



Answer :

Let's solve the given problems step by step.

### Problem 1: Evaluate [tex]\((h + k)(2)\)[/tex]

We are given the functions:
[tex]\[ h(x) = x^2 + 1 \][/tex]
[tex]\[ k(x) = x - 2 \][/tex]

First, we need to evaluate [tex]\( h(2) \)[/tex] and [tex]\( k(2) \)[/tex]:

1. Calculate [tex]\( h(2) \)[/tex]:
[tex]\[ h(2) = 2^2 + 1 = 4 + 1 = 5 \][/tex]

2. Calculate [tex]\( k(2) \)[/tex]:
[tex]\[ k(2) = 2 - 2 = 0 \][/tex]

Now, find [tex]\((h + k)(2)\)[/tex] by adding [tex]\( h(2) \)[/tex] and [tex]\( k(2) \)[/tex]:
[tex]\[ (h+k)(2) = h(2) + k(2) = 5 + 0 = 5 \][/tex]

So,
[tex]\[ (h + k)(2) = 5 \][/tex]

### Problem 2: Evaluate [tex]\((h - k)(3)\)[/tex]

Next, we need to evaluate [tex]\( h(3) \)[/tex] and [tex]\( k(3) \)[/tex]:

1. Calculate [tex]\( h(3) \)[/tex]:
[tex]\[ h(3) = 3^2 + 1 = 9 + 1 = 10 \][/tex]

2. Calculate [tex]\( k(3) \)[/tex]:
[tex]\[ k(3) = 3 - 2 = 1 \][/tex]

Now, find [tex]\((h - k)(3)\)[/tex] by subtracting [tex]\( k(3) \)[/tex] from [tex]\( h(3) \)[/tex]:
[tex]\[ (h-k)(3) = h(3) - k(3) = 10 - 1 = 9 \][/tex]

So,
[tex]\[ (h - k)(3) = 9 \][/tex]

### Problem 3: Evaluate [tex]\( 3h(2) + 2k(3) \)[/tex]

We have already calculated [tex]\( h(2) \)[/tex] and [tex]\( k(2) \)[/tex] and [tex]\( k(3) \)[/tex]:

1. From Problem 1:
[tex]\[ h(2) = 5 \][/tex]
2. From Problem 2:
[tex]\[ k(3) = 1 \][/tex]

Now, calculate:
[tex]\[ 3h(2) + 2k(3) \][/tex]

Substitute the values:
[tex]\[ 3 \times h(2) + 2 \times k(3) = 3 \times 5 + 2 \times 1 = 15 + 2 = 17 \][/tex]

So,
[tex]\[ 3h(2) + 2k(3) = 17 \][/tex]

To summarize:
1. [tex]\((h + k)(2) = 5\)[/tex]
2. [tex]\((h - k)(3) = 9\)[/tex]
3. [tex]\(3h(2) + 2k(3) = 17\)[/tex]