Let’s analyze the problem in the uploaded image.

The problem involves calculating the difference quotient for the given function $f(x) = x^3 - 10x$. The difference quotient formula is:

$\frac{f(2 + h) - f(2)}{h}$

You are provided specific values of $h$ (e.g., $h = 0.1, 0.01, -0.01, -0.1$) and need to calculate the difference quotient for each case.

Steps to Solve:

1. Find $f(2)$: Substitute $x = 2$ into $f(x) = x^3 - 10x$: $f(2) = 2^3 - 10(2) = 8 - 20 = -12.$

2. Find $f(2 + h)$: Substitute $x = 2 + h$ into $f(x) = x^3 - 10x$: $f(2 + h) = (2 + h)^3 - 10(2 + h).$ Expand $(2 + h)^3 = 8 + 12h + 6h^2 + h^3$: $f(2 + h) = 8 + 12h + 6h^2 + h^3 - 20 - 10h = -12 + 2h + 6h^2 + h^3.$

3. Calculate the difference quotient: Substitute $f(2 + h)$ and $f(2)$ into the formula: $\frac{f(2 + h) - f(2)}{h} = \frac{-12 + 2h + 6h^2 + h^3 - (-12)}{h}.$ Simplify the numerator: $\frac{-12 + 2h + 6h^2 + h^3 + 12}{h} = \frac{2h + 6h^2 + h^3}{h}.$ Factor $h$ out of the numerator: $\frac{h(2 + 6h + h^2)}{h} = 2 + 6h + h^2.$

4. Evaluate for each $h$: Substitute the given values of $h$:

   • For $h = 0.1$: $2 + 6(0.1) + (0.1)^2 = 2 + 0.6 + 0.01 = 2.61$.
   • For $h = 0.01$: $2 + 6(0.01) + (0.01)^2 = 2 + 0.06 + 0.0001 = 2.0601$.
   • For $h = -0.01$: $2 + 6(-0.01) + (-0.01)^2 = 2 - 0.06 + 0.0001 = 1.9401$.
   • For $h = -0.1$: $2 + 6(-0.1) + (-0.1)^2 = 2 - 0.6 + 0.01 = 1.41$.

Final Results:

• For $h = 0.1$: $2.61$
• For $h = 0.01$: $2.0601$
• For $h = -0.01$: $1.9401$
• For $h = -0.1$: $1.41$

Let me know if you’d like further clarification!

Related Questions:

1. What happens to the difference quotient as $h \to 0$?
2. How does the difference quotient relate to the derivative of a function?
3. Can this process be generalized for any polynomial $f(x)$?
4. How would the result change if $f(x)$ had an additional constant term?
5. Why is expanding $(2 + h)^3$ important in this problem?

Tip:

Always double-check your algebraic expansions and substitutions when working with polynomials to avoid small errors that propagate through the calculations!