Unit 2 practice test: Differentiation: Definition and Fundamental Properties

$\frac{f(3) - f(1)}{3 - 1}$$= \frac{3.5 - 1.5}{2}$$= 1$.

• Correct: $1$: Change in $y$, 2, over change in $x$, 2.
• $2$: That is the change in $y$ alone; divide by the change in $x$.
• $\frac{1}{2}$: That is run over rise, upside down.
• $\frac{7}{6}$: That is $\frac{3.5}{3}$, a height over an input, not a change over a change.

The slopes are $\frac{h}{2}$, and $\lim_{h \to 0} \frac{h}{2} = 0$. The graph is at its lowest point at $x = 1$, where it is momentarily level.

• Correct: $0$: The limit of $\frac{h}{2}$ as $h \to 0$.
• $1$: That is the slope of the drawn secant line, with $h = 2$; the slopes change as $h$ shrinks.
• $1.5$: That is $f(1)$, the height of the point, not a rate.
• They approach no value, because $h = 0$ gives $\frac{0}{0}$.: $\frac{0}{0}$ appears only if you substitute before simplifying; the simplified quotient has a limit.

The instantaneous rate is the limit of average rates on $[3, 3 + h]$ as $h \to 0$.

• Correct: $\lim_{h \to 0} \frac{f(3 + h) - f(3)}{h}$: The limit of the difference quotient at $a = 3$.
• $\lim_{h \to 0} \left( f(3 + h) - f(3) \right)$: Without dividing by $h$ this is a change in $f$, and its limit is 0 for a continuous $f$.
• $\frac{f(3 + h) - f(3)}{h}$ with $h = 0$: Setting $h = 0$ divides by zero; the rate is a limit, not a value at $h = 0$.
• $\lim_{h \to 0} \frac{f(3 + h)}{h}$: The numerator must be a change in $f$, $f(3 + h) - f(3)$, not a single value.

With $f(4) = 2$, the definition $\lim_{x \to 4} \frac{f(x) - f(4)}{x - 4}$ becomes this limit.

• Correct: $\lim_{x \to 4} \frac{\sqrt{x} - 2}{x - 4}$: The difference quotient in its $x \to a$ form.
• $\lim_{x \to 4} \left( \sqrt{x} - 2 \right)$: The change in height is not divided by the change in $x$; this limit is 0.
• $\lim_{x \to 4} \frac{\sqrt{x}}{x - 4}$: The numerator needs the change $\sqrt{x} - 2$, not the height $\sqrt{x}$; this limit does not exist.
• $\lim_{h \to 0} \frac{\sqrt{4 + h} - 2}{4 + h}$: The denominator must be the change in input, $h$, not the input $4 + h$.

Point $(4, 2)$, slope $f'(4) = \frac{1}{4}$. At $x = 0$ it gives $y = 1$, matching the graph.

• Correct: $y = 2 + \frac{1}{4}(x - 4)$: Height $f(4) = 2$ and slope $f'(4) = \frac{1}{4}$.
• $y = \frac{1}{4} + 2(x - 4)$: The height and the slope are swapped.
• $y = 2 + \frac{1}{4}(x + 4)$: Point-slope form uses $x - 4$.
• $y = 2 + 4(x - 4)$: The slope is $\frac{1}{4}$, not 4; the line in the graph is gentle.

Near the point of tangency the curve and its tangent line become hard to tell apart; that is why the tangent slope is the rate of change at that point.

• Correct: More and more like the tangent line $y = 2 + \frac{1}{4}(x - 4)$: The tangent line is the line the graph looks like up close.
• More and more curved, like a sharper bend: Zooming in straightens a smooth curve; it does not bend it more.
• More and more like the level line $y = 2$: The graph rises with slope $\frac{1}{4}$ there, so it does not look level.
• More and more like the line through the origin and $(4, 2)$: That line has slope $\frac{1}{2}$; the curve near $(4, 2)$ has slope $\frac{1}{4}$.