Math Problem Statement

Evaluate the difference quotient for the given function. Simplify your answer.

f(x) = 4 + 3x − x2,    

f(4 + h) − f(4)

h

Solution

The difference quotient is given by:

f(4+h)f(4)h\frac{f(4 + h) - f(4)}{h}

Given the function f(x)=4+3xx2f(x) = 4 + 3x - x^2, we'll first find f(4+h)f(4 + h) and f(4)f(4).

Step 1: Calculate f(4+h)f(4 + h)

Substitute 4+h4 + h into the function f(x)=4+3xx2f(x) = 4 + 3x - x^2:

f(4+h)=4+3(4+h)(4+h)2f(4 + h) = 4 + 3(4 + h) - (4 + h)^2

First, simplify 3(4+h)3(4 + h):

3(4+h)=12+3h3(4 + h) = 12 + 3h

Next, expand (4+h)2(4 + h)^2:

(4+h)2=16+8h+h2(4 + h)^2 = 16 + 8h + h^2

So, the expression for f(4+h)f(4 + h) becomes:

f(4+h)=4+12+3h(16+8h+h2)f(4 + h) = 4 + 12 + 3h - (16 + 8h + h^2)

Simplify the expression:

f(4+h)=16+3h168hh2=5hh2f(4 + h) = 16 + 3h - 16 - 8h - h^2 = -5h - h^2

Step 2: Calculate f(4)f(4)

Substitute x=4x = 4 into the function:

f(4)=4+3(4)42f(4) = 4 + 3(4) - 4^2

Simplify:

f(4)=4+1216=0f(4) = 4 + 12 - 16 = 0

Step 3: Compute the Difference Quotient

Now, substitute f(4+h)f(4 + h) and f(4)f(4) into the difference quotient formula:

f(4+h)f(4)h=5hh20h\frac{f(4 + h) - f(4)}{h} = \frac{-5h - h^2 - 0}{h}

Simplify the expression:

5hh2h=5h\frac{-5h - h^2}{h} = -5 - h

Thus, the simplified difference quotient is:

5h-5 - h

Do you have any more questions or need further clarification on any steps?

Here are 5 related questions:

  1. What is the difference quotient for f(x)=x23x+2f(x) = x^2 - 3x + 2?
  2. How would the difference quotient change if f(x)f(x) were a linear function?
  3. What is the geometric interpretation of the difference quotient?
  4. How does the difference quotient relate to the derivative of a function?
  5. Evaluate the difference quotient for f(x)=1xf(x) = \frac{1}{x} at a different point.

Tip: The difference quotient is a foundational concept in calculus, leading directly to the definition of the derivative. Practice simplifying these expressions to build your skills!

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Math Problem Analysis

Mathematical Concepts

Difference Quotient
Polynomial Functions
Function Evaluation

Formulas

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Theorems

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Suitable Grade Level

Grades 11-12