Average rate of change, difference quotients, and limits

A car’s position \(s\) is given by the table below.

Find the average velocity over \(0 \le t \le 0.2\text{,}\) then estimate the (instantaneous) velocity at \(t = 0.2\text{.}\)

Average velocity on \([0, 0.2]\text{:}\)

\begin{equation*} \frac{s(0.2) - s(0)}{0.2 - 0} = \frac{0.4 - 0}{0.2} = 2 \text{ ft/s} \end{equation*}

To estimate the velocity at \(t = 0.2\text{,}\) use a centered difference with the neighboring data point at \(t = 0.4\text{:}\)

\begin{equation*} \frac{s(0.4) - s(0)}{0.4 - 0} = \frac{1.4 - 0}{0.4} = 3.5 \text{ ft/s} \end{equation*}

Calories used per minute, by body weight, for three activities:

1. Calories used by a 170 lb person walking for a half-hour?
2. Who uses more: a 120 lb person swimming for 1 hour, or a 200 lb person bicycling for a half-hour?
3. Does walking-calorie-use increase or decrease as weight increases?

(a) A 170 lb person walking burns 4.6 cal/min. Half an hour is 30 minutes:

\begin{equation*} 4.6 \times 30 = 138 \text{ calories} \end{equation*}

(b) 120 lb swimming, 1 hour (60 min): \(6.9 \times 60 = 414\) cal. 200 lb bicycling, 30 min: \(10.8 \times 30 = 324\) cal. The swimmer uses more.

(c) Reading across the Walking row, the values increase as weight increases: Increase.

Let \(f(x) = x^2 + 4\text{.}\) Find the average rate of change:

1. between the points \((-1, f(-1))\) and \((2, f(2))\)
2. between the points \((a, f(a))\) and \((b, f(b))\)

(a) \(f(-1) = 1 + 4 = 5\text{,}\) \(f(2) = 4 + 4 = 8\)

\begin{equation*} \frac{f(2) - f(-1)}{2 - (-1)} = \frac{8 - 5}{3} = \frac{3}{3} = 1 \end{equation*}

(b) In general,

\begin{equation*} \frac{f(b) - f(a)}{b - a} = \frac{(b^2+4) - (a^2+4)}{b-a} = \frac{b^2 - a^2}{b-a} = a + b \end{equation*}

Let \(f(x) = -5x - 6\text{.}\) Find the average rate of change:

1. between the points \((-1, f(-1))\) and \((4, f(4))\)
2. between the points \((a, f(a))\) and \((b, f(b))\)

(a) \(f(-1) = 5 - 6 = -1\text{,}\) \(f(4) = -20 - 6 = -26\)

\begin{equation*} \frac{f(4) - f(-1)}{4 - (-1)} = \frac{-26 - (-1)}{5} = \frac{-25}{5} = -5 \end{equation*}

(b) For any linear function \(f(x) = mx + k\text{,}\) the average rate of change between any two points equals the slope:

\begin{equation*} \frac{f(b) - f(a)}{b - a} = -5 \end{equation*}

Use algebra to evaluate the limit:

Expand the numerator:

\begin{equation*} (3+h)^2 - 9 = 9 + 6h + h^2 - 9 = 6h + h^2 = h(6 + h) \end{equation*}

So

\begin{equation*} \lim_{h \to 0} \frac{h(6+h)}{h} = \lim_{h \to 0} (6 + h) = 6 \end{equation*}

Evaluate

\(f(2+h) = -7(2+h) - 5 = -14 - 7h - 5 = -19 - 7h\text{,}\) and \(f(2) = -19\text{.}\)

\begin{equation*} \frac{f(2+h) - f(2)}{h} = \frac{(-19 - 7h) - (-19)}{h} = \frac{-7h}{h} = -7 \end{equation*}

So the limit is \(-7\) — as expected, since \(f\) is linear with slope \(-7\text{.}\)

Evaluate the limit:

If the limit does not exist, enter DNE.

The function is a ratio of polynomials, and the denominator is nonzero at \(x = 5\) (since \(4 + 5^4 = 629 \ne 0\)), so we can substitute directly:

\begin{equation*} \lim_{x \to 5} \frac{8x}{4 + x^4} = \frac{8(5)}{4 + 5^4} = \frac{40}{629} \end{equation*}

Let \(f(x) = x^2 + 4\) and \(g(x) = 3x - 4\text{.}\)

Find \(f(g(x))\) and \(g(f(x))\text{,}\) and give the domain and range of each.

\begin{equation*} f(g(x)) = (3x-4)^2 + 4 = 9x^2 - 24x + 16 + 4 = 9x^2 - 24x + 20 \end{equation*}

Domain: all real numbers. Range: \(y \ge 4\) (minimum at the vertex of this upward parabola).

\begin{equation*} g(f(x)) = 3(x^2 + 4) - 4 = 3x^2 + 12 - 4 = 3x^2 + 8 \end{equation*}

Domain: all real numbers. Range: \(y \ge 8\text{.}\)