For the function [tex]\( f(x) = -3x^2 - x + 5 \)[/tex], evaluate and fully simplify each of the following:

1. [tex]\( f(x+h) = \square \)[/tex]
2. [tex]\( \frac{f(x+h) - f(x)}{h} = \square \)[/tex]



Answer :

Certainly! Let's solve the problem step-by-step.

### Step 1: Evaluate [tex]\( f(x + h) \)[/tex]

Given the function:
[tex]\[ f(x) = -3x^2 - x + 5 \][/tex]

We need to find [tex]\( f(x + h) \)[/tex]. To do this, we substitute [tex]\( x \)[/tex] with [tex]\( x + h \)[/tex] in the function:

[tex]\[ f(x + h) = -3(x + h)^2 - (x + h) + 5 \][/tex]

Now we expand and simplify:

1. Expand [tex]\((x + h)^2\)[/tex]:
[tex]\[ (x + h)^2 = x^2 + 2xh + h^2 \][/tex]

2. Substitute back into the function:
[tex]\[ f(x + h) = -3(x^2 + 2xh + h^2) - (x + h) + 5 \][/tex]

3. Distribute the -3:
[tex]\[ f(x + h) = -3x^2 - 6xh - 3h^2 - x - h + 5 \][/tex]

Therefore:
[tex]\[ f(x + h) = -3x^2 - 6xh - 3h^2 - x - h + 5 \][/tex]

Simplifying we get:
[tex]\[ f(x+h) = -3(h + x)^2 - h - x + 5 \][/tex]

### Step 2: Evaluate the difference quotient and simplify

The difference quotient formula is:
[tex]\[ \frac{f(x + h) - f(x)}{h} \][/tex]

We already have [tex]\( f(x + h) \)[/tex] and [tex]\( f(x) \)[/tex]:

[tex]\[ f(x + h) = -3x^2 - 6xh - 3h^2 - x - h + 5 \][/tex]
[tex]\[ f(x) = -3x^2 - x + 5 \][/tex]

Next, we calculate [tex]\( f(x + h) - f(x) \)[/tex]:

[tex]\[ f(x + h) - f(x) = (-3x^2 - 6xh - 3h^2 - x - h + 5) - (-3x^2 - x + 5) \][/tex]

Distribute and combine like terms:

[tex]\[ f(x + h) - f(x) = -3x^2 - 6xh - 3h^2 - x - h + 5 + 3x^2 + x - 5 \][/tex]

[tex]\[ f(x + h) - f(x) = - 6xh - 3h^2 - h \][/tex]

Now, divide by [tex]\( h \)[/tex]:

[tex]\[ \frac{f(x + h) - f(x)}{h} = \frac{-6xh - 3h^2 - h}{h} \][/tex]

Simplify by dividing each term in the numerator by [tex]\( h \)[/tex]:

[tex]\[ \frac{f(x + h) - f(x)}{h} = \frac{-6xh}{h} + \frac{-3h^2}{h} + \frac{-h}{h} \][/tex]

[tex]\[ \frac{f(x + h) - f(x)}{h} = -6x - 3h - 1 \][/tex]

So, the simplified form of the difference quotient is:

[tex]\[ \boxed{-6x - 3h - 1} \][/tex]