Question Statement

Find the average rate of change of the function $f(x)=x^3+2x^2-4x$ over the interval $[-1,2]$.

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Background and Explanation

The average rate of change of a function over an interval $[a,b]$ represents the slope of the secant line connecting the endpoints, calculated as $\frac{\Delta y}{\Delta x}=\frac{f(b)-f(a)}{b-a}$. This requires evaluating the difference quotient by expanding $f(x+\Delta x)$ and simplifying.

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Solution

Let $y=f(x)=x^3+2x^2-4x$

The average rate of change in $y$ is given by the difference quotient:

$\frac{\Delta y}{\Delta x}=\frac{f(x+\Delta x)-f(x)}{\Delta x}$

First, expand $f(x+\Delta x)$: $f(x+\Delta x)=(x+\Delta x{)}^3+2(x+\Delta x{)}^2-4(x+\Delta x)$

Expanding each term:

• $(x+\Delta x{)}^3=x^3+3x^2\Delta x+3x\Delta x^2+\Delta x^3$
• $2(x+\Delta x{)}^2=2(x^2+2x\Delta x+\Delta x^2)=2x^2+4x\Delta x+2\Delta x^2$
• $-4(x+\Delta x)=-4x-4\Delta x$

Substituting back: $f(x+\Delta x)=x^3+3x^2\Delta x+3x\Delta x^2+\Delta x^3+2x^2+4x\Delta x+2\Delta x^2-4x-4\Delta x$

Now compute the numerator $f(x+\Delta x)-f(x)$: $\begin{array}{rl}& =\left[x^3+3x^2\Delta x+3x\Delta x^2+\Delta x^3+2x^2+4x\Delta x+2\Delta x^2-4x-4\Delta x\right]\\ & -\left[x^3+2x^2-4x\right]\end{array}$

Canceling like terms ($x^3$, $2x^2$, and $-4x$): $=3x^2\Delta x+3x\Delta x^2+\Delta x^3+4x\Delta x+2\Delta x^2-4\Delta x$

Factor out $\Delta x$: $=\Delta x\left[\Delta x^2+3x^2+3x\Delta x+2\Delta x+4x-4\right]$

Dividing by $\Delta x$: $\frac{\Delta y}{\Delta x}=\Delta x^2+3x^2+3x\Delta x+2\Delta x+4x-4$

Given the interval $[-1,2]$:

• Let $x=-1$ (the starting point)
• $\Delta x=2-(-1)=3$

Substitute $x=-1$ and $\Delta x=3$ into equation (1):

$\begin{array}{rl}\frac{\Delta y}{\Delta x}& =(3{)}^2+3(-1{)}^2+3(-1)(3)+2(3)+4(-1)-4\\ & =9+3-9+6-4-4\\ & =1\end{array}$

Thus, the average rate of change of $f(x)$ over the interval $[-1,2]$ is $\boxed{1}$.

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Key Formulas or Methods Used

• Difference Quotient: $\frac{\Delta y}{\Delta x}=\frac{f(x+\Delta x)-f(x)}{\Delta x}$
• Binomial Expansion: $(x+\Delta x{)}^3=x^3+3x^2\Delta x+3x\Delta x^2+\Delta x^3$
• Perfect Square: $(x+\Delta x{)}^2=x^2+2x\Delta x+\Delta x^2$
• Interval Parameters: For $[a,b]$, use $x=a$ and $\Delta x=b-a$

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Summary of Steps

1. Set up the difference quotient formula for average rate of change
2. Expand $f(x+\Delta x)$ using binomial theorem for the cubic and quadratic terms
3. Subtract $f(x)$ from the expanded $f(x+\Delta x)$ and collect like terms
4. Factor out $\Delta x$ from the numerator and simplify the quotient
5. Identify $x=-1$ and $\Delta x=3$ corresponding to the interval $[-1,2]$
6. Substitute these values into the simplified rate expression and evaluate to get $1$