Let f(x) = 2х^2 - x.

Evaluate and simplify. f(x+h)- f(x)/h a: 4x + 2h-1 b:4x-2h-1-2x/h c:4x+2h+1 d:4-2h+1-2x/h e:2h-1 f:2h-1-2x/h

To evaluate and simplify \(\frac{f(x+h) - f(x)}{h}\) for the function \(f(x) = 2x^2 - x\), let's follow these steps:

1. **Find \(f(x+h)\):**

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

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

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

   Therefore,

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

2. **Find \(f(x+h) - f(x)\):**

   Now, subtract \(f(x)\) from \(f(x+h)\):

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

   Simplifying:

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

3. **Divide by \(h\):**

   Finally, divide by \(h\):

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

   Simplifying by dividing each term by \(h\):

   \[
   \frac{f(x+h) - f(x)}{h} = 4x + 2h - 1
   \]

The correct answer is \( \boxed{4x + 2h - 1} \).

Would you like more details on this, or do you have any other questions?

Here are five related questions:
1. How do you find the derivative of \( f(x) \) using the limit definition?
2. Can you apply the same method to find the difference quotient for \( f(x) = 3x^2 - 4x + 1 \)?
3. What is the significance of the difference quotient in calculus?
4. How would you interpret the result \( 4x + 2h - 1 \) in terms of a slope?
5. How does the expression change if \( h \) approaches zero?

**Tip:** Understanding the difference quotient is essential, as it forms the basis for finding derivatives in calculus.

Learn how to evaluate and simplify the difference quotient for the quadratic function f(x) = 2x^2 - x. Follow step-by-step calculations to understand the concept of the difference quotient. Suitable for advanced high school students studying calculus and quadratic functions.

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