https://file.aiquickdraw.com/mathbot/file/1726104383926ivkx11vr.jpg

We are tasked with calculating and simplifying the difference quotient for the function \( f(x) = x^5 \). The difference quotient formula is:

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

Substituting \( f(x) = x^5 \), we get:

\[
\frac{(x+h)^5 - x^5}{h}
\]

### Step-by-Step Solution:

1. **Expand \( (x+h)^5 \) using the binomial theorem:**

\[
(x+h)^5 = x^5 + 5x^4h + 10x^3h^2 + 10x^2h^3 + 5xh^4 + h^5
\]

2. **Substitute the expansion back into the difference quotient:**

\[
\frac{x^5 + 5x^4h + 10x^3h^2 + 10x^2h^3 + 5xh^4 + h^5 - x^5}{h}
\]

3. **Simplify the expression by canceling out \( x^5 \):**

\[
\frac{5x^4h + 10x^3h^2 + 10x^2h^3 + 5xh^4 + h^5}{h}
\]

4. **Factor out \( h \) from the numerator:**

\[
\frac{h(5x^4 + 10x^3h + 10x^2h^2 + 5xh^3 + h^4)}{h}
\]

5. **Cancel the \( h \) in the numerator and denominator (assuming \( h \neq 0 \)):**

\[
5x^4 + 10x^3h + 10x^2h^2 + 5xh^3 + h^4
\]

This is the simplified form of the difference quotient for \( f(x) = x^5 \). The hint that the fraction reduces into a polynomial expression is confirmed, as the final result is a polynomial in terms of \( h \).

Would you like more details on any step? Here are five related questions:

1. How does the binomial theorem help in expanding expressions like \( (x + h)^5 \)?
2. What is the significance of the difference quotient in calculus?
3. How does this process relate to finding the derivative of \( f(x) = x^5 \)?
4. What happens to the terms involving \( h \) as \( h \to 0 \)?
5. How would this process change for a different function, such as \( f(x) = x^3 \)?

**Tip:** The difference quotient is a critical concept as it lays the foundation for understanding derivatives and rates of change in calculus.

Calculate and simplify the Difference Quotient (f(x+h)-f(x))/h for f(x) = x^5.

Learn how to calculate and simplify the Difference Quotient for the function f(x) = x^5. This problem covers polynomial expansion using the binomial theorem and key calculus concepts like the difference quotient. Suitable for students in Grades 11-12 or introductory Calculus.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation