Using the definition of the​ derivative, find f prime left parenthesis x right parenthesis
. 
Then find f prime left parenthesis 1 right parenthesis
​, 
f prime left parenthesis 2 right parenthesis
​, 
and f prime left parenthesis 3 right parenthesis
 
when the derivative exists.
​f(x)equals
negative x squared plus 4 x minus 8
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Part 1
f prime left parenthesis x right parenthesis
equalsenter your response here
​(Type an expression using x as the​ variable.)

To solve this problem using the definition of the derivative, we'll proceed with the following steps.

### Step 1: Definition of the Derivative
The definition of the derivative \( f'(x) \) at a point \( x \) is given by:

\[
f'(x) = \lim_{{h \to 0}} \frac{f(x+h) - f(x)}{h}
\]

Given the function \( f(x) = -x^2 + 4x - 8 \), we need to use this definition to find the derivative.

### Step 2: Find \( f(x+h) \)
First, substitute \( x + h \) into the function:

\[
f(x+h) = -(x+h)^2 + 4(x+h) - 8
\]

Expand \( (x+h)^2 \):

\[
(x+h)^2 = x^2 + 2xh + h^2
\]

So:

\[
f(x+h) = -(x^2 + 2xh + h^2) + 4(x + h) - 8
\]

Expanding further:

\[
f(x+h) = -x^2 - 2xh - h^2 + 4x + 4h - 8
\]

### Step 3: Find \( f(x+h) - f(x) \)
Now, subtract \( f(x) \) from \( f(x+h) \):

\[
f(x+h) - f(x) = (-x^2 - 2xh - h^2 + 4x + 4h - 8) - (-x^2 + 4x - 8)
\]

Simplify:

\[
f(x+h) - f(x) = -2xh - h^2 + 4h
\]

### Step 4: Divide by \( h \)
Now, divide by \( h \):

\[
\frac{f(x+h) - f(x)}{h} = \frac{-2xh - h^2 + 4h}{h}
\]

Factor \( h \) out of the numerator:

\[
\frac{f(x+h) - f(x)}{h} = \frac{h(-2x - h + 4)}{h}
\]

Cancel \( h \):

\[
\frac{f(x+h) - f(x)}{h} = -2x - h + 4
\]

### Step 5: Take the Limit as \( h \to 0 \)
Take the limit as \( h \) approaches 0:

\[
f'(x) = \lim_{{h \to 0}} (-2x - h + 4) = -2x + 4
\]

### Final Result
The derivative \( f'(x) \) is:

\[
f'(x) = -2x + 4
\]

### Step 6: Find \( f'(1) \), \( f'(2) \), and \( f'(3) \)

1. \( f'(1) = -2(1) + 4 = -2 + 4 = 2 \)
2. \( f'(2) = -2(2) + 4 = -4 + 4 = 0 \)
3. \( f'(3) = -2(3) + 4 = -6 + 4 = -2 \)

### Summary
- \( f'(x) = -2x + 4 \)
- \( f'(1) = 2 \)
- \( f'(2) = 0 \)
- \( f'(3) = -2 \)

Would you like me to elaborate on any of these steps, or do you have any questions?

### Related Questions
1. What is the geometric interpretation of the derivative \( f'(x) \) at a given point?
2. How can we use the derivative to find the local maxima or minima of the function?
3. What does it mean if the derivative is equal to zero at a point?
4. How would you find the second derivative of the function \( f(x) \)?
5. How does the derivative relate to the slope of the tangent line at a point?

### Tip
The derivative represents the rate of change or the slope of the tangent line to the curve at a particular point.

Using the definition of the derivative, find f'(x). Then find f'(1), f'(2), and f'(3) when the derivative exists. f(x) = -x^2 + 4x - 8.

Learn how to apply the definition of the derivative to find the derivative of the quadratic function f(x) = -x^2 + 4x - 8. Includes step-by-step calculation of f'(1), f'(2), and f'(3), demonstrating key calculus concepts like limits and differentiation. Suitable for students in Grades 10-12.

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