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)equalsnegative x squared plus 5 x minus 9
To find the​ derivative, complete the limit as h approaches 0 for StartFraction f left parenthesis x plus h right parenthesis minus f left parenthesis x right parenthesis Over h EndFraction
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To find the derivative \( f'(x) \) of the function \( f(x) = -x^2 + 5x - 9 \) using the definition of the derivative, we need to compute the following limit:

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

### Step 1: Compute \( f(x + h) \)
Substitute \( x + h \) into the function \( f(x) \):

\[
f(x + h) = -(x + h)^2 + 5(x + h) - 9
\]
Expand this expression:
\[
f(x + h) = -\left( x^2 + 2xh + h^2 \right) + 5x + 5h - 9
\]
Simplifying:
\[
f(x + h) = -x^2 - 2xh - h^2 + 5x + 5h - 9
\]

### Step 2: Compute \( f(x + h) - f(x) \)
Now subtract \( f(x) = -x^2 + 5x - 9 \) from \( f(x + h) \):
\[
f(x + h) - f(x) = \left( -x^2 - 2xh - h^2 + 5x + 5h - 9 \right) - \left( -x^2 + 5x - 9 \right)
\]
Distribute the negative sign:
\[
f(x + h) - f(x) = -x^2 - 2xh - h^2 + 5x + 5h - 9 + x^2 - 5x + 9
\]
Simplify the terms:
\[
f(x + h) - f(x) = -2xh - h^2 + 5h
\]

### Step 3: Form the difference quotient
Now, divide by \( h \):
\[
\frac{f(x + h) - f(x)}{h} = \frac{-2xh - h^2 + 5h}{h}
\]
Simplify the expression:
\[
\frac{f(x + h) - f(x)}{h} = -2x - h + 5
\]

### Step 4: Take the limit as \( h \to 0 \)
Now take the limit as \( h \) approaches 0:
\[
f'(x) = \lim_{{h \to 0}} (-2x - h + 5) = -2x + 5
\]

Thus, the derivative of \( f(x) \) is:
\[
f'(x) = -2x + 5
\]

### Step 5: Find \( f'(1) \), \( f'(2) \), and \( f'(3) \)
- \( f'(1) = -2(1) + 5 = -2 + 5 = 3 \)
- \( f'(2) = -2(2) + 5 = -4 + 5 = 1 \)
- \( f'(3) = -2(3) + 5 = -6 + 5 = -1 \)

### Summary:
- \( f'(x) = -2x + 5 \)
- \( f'(1) = 3 \)
- \( f'(2) = 1 \)
- \( f'(3) = -1 \)

Would you like more details on any of these steps or further explanations?

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**Related Questions:**
1. What is the geometric meaning of the derivative \( f'(x) \) at a point?
2. How would the second derivative of \( f(x) = -x^2 + 5x - 9 \) be computed?
3. What is the interpretation of the values \( f'(1) \), \( f'(2) \), and \( f'(3) \)?
4. Can we use the power rule to differentiate \( f(x) = -x^2 + 5x - 9 \) and get the same result?
5. How would the derivative change if the function were \( f(x) = -x^3 + 5x - 9 \)?

**Tip:** The definition of the derivative is a fundamental concept in calculus that captures the rate of change of a function at any point. It’s crucial for understanding motion, optimization, and various other mathematical applications.

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 + 5x - 9. To find the derivative, complete the limit as h approaches 0 for (f(x + h) - f(x)) / h.

Learn how to find the derivative of the quadratic function f(x) = -x^2 + 5x - 9 using the limit definition of the derivative. This step-by-step guide covers key calculus concepts such as limits, derivatives, and their applications. Includes examples of f'(1), f'(2), and f'(3). Suitable for high school and college calculus students.

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