Math Problem Statement

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 .

Solution

To find the derivative f(x)f'(x) of the function f(x)=x2+5x9f(x) = -x^2 + 5x - 9 using the definition of the derivative, we need to compute the following limit:

f(x)=limh0f(x+h)f(x)hf'(x) = \lim_{{h \to 0}} \frac{{f(x + h) - f(x)}}{h}

Step 1: Compute f(x+h)f(x + h)

Substitute x+hx + h into the function f(x)f(x):

f(x+h)=(x+h)2+5(x+h)9f(x + h) = -(x + h)^2 + 5(x + h) - 9 Expand this expression: f(x+h)=(x2+2xh+h2)+5x+5h9f(x + h) = -\left( x^2 + 2xh + h^2 \right) + 5x + 5h - 9 Simplifying: f(x+h)=x22xhh2+5x+5h9f(x + h) = -x^2 - 2xh - h^2 + 5x + 5h - 9

Step 2: Compute f(x+h)f(x)f(x + h) - f(x)

Now subtract f(x)=x2+5x9f(x) = -x^2 + 5x - 9 from f(x+h)f(x + h): f(x+h)f(x)=(x22xhh2+5x+5h9)(x2+5x9)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)=x22xhh2+5x+5h9+x25x+9f(x + h) - f(x) = -x^2 - 2xh - h^2 + 5x + 5h - 9 + x^2 - 5x + 9 Simplify the terms: f(x+h)f(x)=2xhh2+5hf(x + h) - f(x) = -2xh - h^2 + 5h

Step 3: Form the difference quotient

Now, divide by hh: f(x+h)f(x)h=2xhh2+5hh\frac{f(x + h) - f(x)}{h} = \frac{-2xh - h^2 + 5h}{h} Simplify the expression: f(x+h)f(x)h=2xh+5\frac{f(x + h) - f(x)}{h} = -2x - h + 5

Step 4: Take the limit as h0h \to 0

Now take the limit as hh approaches 0: f(x)=limh0(2xh+5)=2x+5f'(x) = \lim_{{h \to 0}} (-2x - h + 5) = -2x + 5

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

Step 5: Find f(1)f'(1), f(2)f'(2), and f(3)f'(3)

  • f(1)=2(1)+5=2+5=3f'(1) = -2(1) + 5 = -2 + 5 = 3
  • f(2)=2(2)+5=4+5=1f'(2) = -2(2) + 5 = -4 + 5 = 1
  • f(3)=2(3)+5=6+5=1f'(3) = -2(3) + 5 = -6 + 5 = -1

Summary:

  • f(x)=2x+5f'(x) = -2x + 5
  • f(1)=3f'(1) = 3
  • f(2)=1f'(2) = 1
  • f(3)=1f'(3) = -1

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


Related Questions:

  1. What is the geometric meaning of the derivative f(x)f'(x) at a point?
  2. How would the second derivative of f(x)=x2+5x9f(x) = -x^2 + 5x - 9 be computed?
  3. What is the interpretation of the values f(1)f'(1), f(2)f'(2), and f(3)f'(3)?
  4. Can we use the power rule to differentiate f(x)=x2+5x9f(x) = -x^2 + 5x - 9 and get the same result?
  5. How would the derivative change if the function were f(x)=x3+5x9f(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.

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Math Problem Analysis

Mathematical Concepts

Calculus
Derivatives
Limits
Quadratic Functions

Formulas

f'(x) = lim(h → 0) [f(x + h) - f(x)] / h
f(x) = -x^2 + 5x - 9
f'(x) = -2x + 5

Theorems

Definition of the Derivative
Limit Theorem

Suitable Grade Level

Grades 11-12, College Introductory Calculus