Question: Find the average rate of change of $g(x)=x^3-2x+1$
1. **State the problem:** We need to find the average rate of change of the function $g(x)=x^3-2x+1$ over an interval $[a,b]$.
2. **Formula:** The average rate of change of a function $g(x)$ from $x=a$ to $x=b$ is given by:
$$\frac{g(b)-g(a)}{b-a}$$
This formula calculates the slope of the secant line connecting the points $(a,g(a))$ and $(b,g(b))$ on the graph.
3. **Choose interval:** Since the problem does not specify an interval, let's find the average rate of change between two points $a$ and $b$.
4. **Calculate $g(a)$ and $g(b)$:**
$$g(a) = a^3 - 2a + 1$$
$$g(b) = b^3 - 2b + 1$$
5. **Substitute into formula:**
$$\frac{g(b)-g(a)}{b-a} = \frac{(b^3 - 2b + 1) - (a^3 - 2a + 1)}{b-a} = \frac{b^3 - 2b + 1 - a^3 + 2a - 1}{b-a}$$
6. **Simplify numerator:**
$$b^3 - a^3 - 2b + 2a + (1 - 1) = b^3 - a^3 - 2b + 2a$$
7. **Rewrite numerator:**
$$b^3 - a^3 = (b - a)(b^2 + ab + a^2)$$
8. **Factor numerator:**
$$ (b - a)(b^2 + ab + a^2) - 2(b - a) = (b - a)(b^2 + ab + a^2 - 2)$$
9. **Divide numerator and denominator:**
$$\frac{(b - a)(b^2 + ab + a^2 - 2)}{b - a}$$
10. **Cancel common factor:**
$$\frac{\cancel{(b - a)}(b^2 + ab + a^2 - 2)}{\cancel{(b - a)}} = b^2 + ab + a^2 - 2$$
**Final answer:** The average rate of change of $g(x)$ from $x=a$ to $x=b$ is:
$$b^2 + ab + a^2 - 2$$
This expression gives the average rate of change for any interval $[a,b]$.