To address this problem thoroughly, let’s work step by step:

The function is given as: $f(x) = x^3 - 4x$ We are tasked with evaluating the difference quotient: $\frac{f(2+h) - f(2)}{h}$ for several small values of $h$ and interpreting the result.

Step 1: Find $f(2)$

Substitute $x = 2$ into $f(x)$: $f(2) = 2^3 - 4(2) = 8 - 8 = 0$

Step 2: Find $f(2+h)$

The function is $f(x) = x^3 - 4x$. To find $f(2+h)$, substitute $x = 2+h$ into the equation: $f(2+h) = (2+h)^3 - 4(2+h)$ Expand $(2+h)^3$ using the binomial theorem: $(2+h)^3 = 8 + 12h + 6h^2 + h^3$ Thus: $f(2+h) = 8 + 12h + 6h^2 + h^3 - 4(2+h)$ Simplify the terms: $f(2+h) = 8 + 12h + 6h^2 + h^3 - 8 - 4h$ Combine like terms: $f(2+h) = 8h + 6h^2 + h^3$

Step 3: Difference Quotient

The difference quotient is: $\frac{f(2+h) - f(2)}{h}$ Since $f(2) = 0$, this simplifies to: $\frac{f(2+h)}{h} = \frac{8h + 6h^2 + h^3}{h}$ Simplify by dividing through by $h$ (assuming $h \neq 0$): $\frac{f(2+h)}{h} = 8 + 6h + h^2$

Step 4: Evaluate for Small $h$

We now calculate the expression $8 + 6h + h^2$ for the given values of $h$:

1. $h = 0.1$: $8 + 6(0.1) + (0.1)^2 = 8 + 0.6 + 0.01 = 8.61$

2. $h = 0.01$: $8 + 6(0.01) + (0.01)^2 = 8 + 0.06 + 0.0001 = 8.0601$

3. $h = -0.01$: $8 + 6(-0.01) + (-0.01)^2 = 8 - 0.06 + 0.0001 = 7.9401$

4. $h = -0.1$: $8 + 6(-0.1) + (-0.1)^2 = 8 - 0.6 + 0.01 = 7.41$

Step 5: Derivative Interpretation

The derivative of $f(x)$ at $x = 2$ is the limit of the difference quotient as $h \to 0$: $f'(2) = \lim_{h \to 0} \frac{f(2+h) - f(2)}{h}$ From the simplified formula $8 + 6h + h^2$, taking the limit as $h \to 0$: $f'(2) = 8$

This matches the expected integer value of $8$.

Final Answer:

The derivative of $f(x) = x^3 - 4x$ at $x = 2$ is $8$.

Do you need further clarifications or additional examples?

Related Questions:

1. How is the difference quotient derived for a given function?
2. What is the formal definition of the derivative using limits?
3. How can you compute derivatives of polynomials directly?
4. What role does the binomial theorem play in simplifying expressions like $(2+h)^3$?
5. How does the difference quotient approximate the derivative for small $h$?

Tip: The derivative represents the slope of the tangent line to the curve at a given point. Approximations become more accurate as $h \to 0$.