```
\begin{array}{l}
h(x)=x^2+1 \quad k(x)=x-2 \\
(h+k)(2)= \square \\
(h-k)(3)= \square \\
\end{array}

Evaluate \, 3h(2)+2k(3)= \square \square.
```



Answer :

Certainly! Let's go through each part step-by-step.

First, the functions [tex]\( h(x) \)[/tex] and [tex]\( k(x) \)[/tex] are defined as follows:
- [tex]\( h(x) = x^2 + 1 \)[/tex]
- [tex]\( k(x) = x - 2 \)[/tex]

### Step 1: Calculate [tex]\((h + k)(2)\)[/tex]
We need to find [tex]\( h(2) \)[/tex] and [tex]\( k(2) \)[/tex] and then add them together.

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]

3. Add [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].

### Step 2: Calculate [tex]\((h - k)(3)\)[/tex]
We need to find [tex]\( h(3) \)[/tex] and [tex]\( k(3) \)[/tex] and then subtract [tex]\( k(3) \)[/tex] from [tex]\( h(3) \)[/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]

3. Subtract [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].

### Step 3: Evaluate [tex]\( 3h(2) + 2k(3) \)[/tex]
We already have computed [tex]\( h(2) \)[/tex] and [tex]\( k(3) \)[/tex]:

1. [tex]\( h(2) = 5 \)[/tex] (from Step 1)
2. [tex]\( k(3) = 1 \)[/tex] (from Step 2)

Now calculate:

1. [tex]\( 3h(2) \)[/tex]:
[tex]\[ 3h(2) = 3 \times 5 = 15 \][/tex]

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

3. Add these results together:
[tex]\[ 3h(2) + 2k(3) = 15 + 2 = 17 \][/tex]

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

### Summary:
- [tex]\((h + k)(2) = 5\)[/tex]
- [tex]\((h - k)(3) = 9\)[/tex]
- [tex]\( 3h(2) + 2k(3) = 17 \)[/tex]