Math Problem Statement

the Difference Quotient for f(x) = -x^2-x+5

Solution

To find the Difference Quotient for the function f(x)=x2x+5f(x) = -x^2 - x + 5, we use the following formula:

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

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

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

f(x+h)=(x+h)2(x+h)+5f(x+h) = -(x+h)^2 - (x+h) + 5 Expanding this expression: f(x+h)=(x2+2xh+h2)(x+h)+5f(x+h) = -(x^2 + 2xh + h^2) - (x + h) + 5 f(x+h)=x22xhh2xh+5f(x+h) = -x^2 - 2xh - h^2 - x - h + 5

Step 2: Difference Quotient

Now, substitute f(x+h)f(x+h) and f(x)f(x) into the Difference Quotient formula: f(x+h)f(x)h=(x22xhh2xh+5)(x2x+5)h\frac{f(x+h) - f(x)}{h} = \frac{(-x^2 - 2xh - h^2 - x - h + 5) - (-x^2 - x + 5)}{h} Simplify the numerator: (x22xhh2xh+5)(x2x+5)(-x^2 - 2xh - h^2 - x - h + 5) - (-x^2 - x + 5) This simplifies to: 2xhh2h-2xh - h^2 - h

Step 3: Simplify the Difference Quotient

Now, divide the simplified expression by hh: 2xhh2hh=2xh1\frac{-2xh - h^2 - h}{h} = -2x - h - 1

Thus, the Difference Quotient for f(x)=x2x+5f(x) = -x^2 - x + 5 is:

2xh1\boxed{-2x - h - 1}

Would you like any further details on this process or any clarifications?


Here are 5 related questions you might find helpful:

  1. How is the Difference Quotient used to find the derivative of a function?
  2. What is the Difference Quotient for a linear function?
  3. Can the Difference Quotient be applied to trigonometric functions?
  4. How does the Difference Quotient change for functions with higher powers of xx?
  5. How can we interpret the result of the Difference Quotient geometrically?

Tip: The Difference Quotient formula is the foundation of the derivative, which gives the slope of the tangent line at any point on a curve.

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Math Problem Analysis

Mathematical Concepts

Algebra
Calculus
Difference Quotient

Formulas

Difference Quotient: (f(x+h) - f(x)) / h

Theorems

Concept of the Difference Quotient as a foundation for the derivative

Suitable Grade Level

Grades 11-12 (Pre-Calculus/Calculus)