Define a tangent line to a curve at a point.

A tangent line at a point is a line that just touches the curve there, without cutting it at or near that point. Its slope is what the derivative of the function measures.

True or False? A tangent line to a curve can cross that same curve at another point.

True. The tangent only has to touch without cutting at or near the point of tangency . Further along the curve it is free to cross, and for many curves it does.

True or False? When a table offers several pairs of values around a point, any pair straddling it gives an equally good estimate of the derivative.

False. The pair that bounds the point most tightly gives the best estimate, because a shorter interval leaves the secant less room to drift away from the tangent. Take the two values closest to the point, one on each side of it.

How do you estimate a derivative from a drawn tangent line?

Draw the tangent at the point, pick two points on the tangent line rather than on the curve, and work out the slope between them. The accuracy depends entirely on how well the tangent was drawn, which is why the result is an estimate.

What must be true of a function before its derivative can be estimated from nearby points on its graph?

It must be continuous and differentiable across the interval containing those points. Without that, the graph could do something between the two points that no secant slope would reveal.

Define a differentiable function .

A differentiable function is one whose derivative exists at every point of its domain. A function with points where the derivative fails can still be made differentiable by restricting its domain so that those points are left out.

Name the four ways a continuous function can fail to be differentiable at a point.

The four are a corner , a cusp , a vertical tangent , and oscillation of the difference quotient near the point. Each is a different way for the limit defining the derivative to fail there.

What does differentiability at a point tell you about continuity there?

If a function is differentiable at a point then it is continuous at that point. In practice the implication is often used the other way round: a function that is not continuous somewhere cannot be differentiable there either.

AP®MathsCollege BoardCalculus BCFlashcardsUnit 2: Differentiation Definition & Fundamental PropertiesDefinition of Differentiation

Definition of Differentiation (College Board AP® Calculus BC): Flashcards

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FrontAverage Rate of Change
Define the average rate of change of a function between two points.

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Cards in this collection (34)

Define the average rate of change of a function between two points.
The average rate of change between two points on the graph is the slope of the line segment joining them.
For points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ it is $\frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ , the change in $y$ divided by the change in $x$ .
True or False?
The average rate of change of $f$ between two points depends only on the values of $f$ at those two points.
True.
It is the slope of the line segment joining them, so nothing the function does in between has any effect on it.
Two functions that agree at the two ends but behave quite differently between them share the same average rate of change there.
A second point lies $h$ units to the right of the point with $x$ -coordinate $a$ . Fill in the two gaps in the difference quotient:
$\frac{f (______) - f (a)}{______}$
The completed difference quotient is $\frac{f (a + h) - f (a)}{h}$ .
The denominator is the horizontal gap between the two points, which here is simply $h$ .
For $f (x) = 3 x^{3} + 2 x - 8$ , find the average rate of change between $x = - 1$ and $x = 2$ .
The average rate of change is $11$ .
Here $f (2) = 20$ and $f (- 1) = - 13$ , so the rate is $\frac{20 - (- 13)}{2 - (- 1)} = \frac{33}{3}$ .
When is the average rate of change between two points undefined?
When the two points share the same $x$ -coordinate, so the change in $x$ is zero and the formula divides by zero.
The line segment joining them is then vertical, and a vertical line has no slope.
Define the instantaneous rate of change of a function.
The instantaneous rate of change of $f$ at $x = a$ is the slope of its graph at that single point, rather than between two points.
It is defined as the limit of the average rate of change as the second point closes in on the first.
Why does the slope of the secant line through $A$ and $B$ approach the instantaneous rate of change at $A$ as $B$ moves towards $A$ ?
Because the shorter the gap between the two points, the more closely the secant matches the direction the curve is heading in at $A$ .
The instantaneous rate is defined as the limit of that slope as the gap shrinks to nothing, so it is exactly what the secant slopes are closing in on.
True or False?
The instantaneous rate of change at $x = a$ can be found by putting $x = a$ into the difference quotient.
False.
That makes the numerator and the denominator both zero, and $\frac{0}{0}$ is not a value.
Taking a limit is what gets round it, describing the quotient as the gap closes without ever setting the gap to zero.
Fill in the two missing limits in the definition of the instantaneous rate of change at $x = a$ :
$\lim_{x \to______} \frac{f (x) - f (a)}{x - a}$
$\lim_{h \to______} \frac{f (a + h) - f (a)}{h}$
The completed forms are $\lim_{x \to a} \frac{f (x) - f (a)}{x - a}$ and $\lim_{h \to 0} \frac{f (a + h) - f (a)}{h}$ .
Each says the same thing in a different labelling: shrink the gap between the two points to nothing.
Use a limit to find the instantaneous rate of change of $f (x) = x^{2}$ at $x = 2$ .
It is $4$ .
Writing $\lim_{x \to 2} \frac{x^{2} - 4}{x - 2}$ and factoring the numerator gives $\lim_{x \to 2} (x + 2)$ , which evaluates to $4$ .
The instantaneous rate of change of $f$ at $x = a$ has another name in calculus. What is it?
It is the derivative of $f$ at $a$ , written $f^{'} (a)$ .
So the two limit forms defining an instantaneous rate of change are also the definition of the derivative at a point.
Define a tangent line to a curve at a point.
A tangent line at a point is a line that just touches the curve there, without cutting it at or near that point.
Its slope is what the derivative of the function measures.
What does $f^{'} (2)$ tell you about the graph of $y = f (x)$ ?
It is the slope of the tangent line to the graph at the point where $x = 2$ .
Equivalently it is the slope of the curve itself at that point.
For $y = f (x)$ , three notations are used for the derivative. Fill in the two that are missing:
$f^{'} (x)$ and $______$ and also $______$ as a third form.
The completed list is $f^{'} (x)$ and $\frac{d y}{d x}$ and also $y^{'}$ as a third form.
All three mean the same thing, and you will also see "the derivative of" written as "differentiated".
What is true of $f^{'} (x)$ where the tangent to the graph of $f$ is horizontal?
It is zero, since a horizontal line has slope zero.
So solving $f^{'} (x) = 0$ picks out exactly the points at which the tangent is horizontal.
True or False?
A tangent line to a curve can cross that same curve at another point.
True.
The tangent only has to touch without cutting at or near the point of tangency.
Further along the curve it is free to cross, and for many curves it does.
What is the equation of the tangent to $y = f (x)$ at the point $(a, b)$ ?
It is $y - b = f^{'} (a) (x - a)$ .
This is the straight line through $(a, b)$ whose slope $m$ has been taken to be the value of the derivative there.
The graph of $f (x) = x^{\frac{1}{3}}$ has a vertical tangent at $x = 0$ , while $h (x) = \frac{1}{x}$ has a vertical asymptote there. What is the difference?
A vertical tangent touches the graph at a point the curve really passes through, and there $f^{'} (0)$ is undefined because a vertical line has no slope.
A vertical asymptote is a line the curve only approaches, at an $x$ -value where the function has no value at all and therefore no tangent either.
For $f (x) = x^{2} - 3 x - 4$ it is known that $f^{'} (4) = 5$ . Find the equation of the tangent to the graph at $x = 4$ .
The tangent is $y = 5 x - 20$ .
Since $f (4) = 16 - 12 - 4 = 0$ the point is $(4, 0)$ , and $y - 0 = 5 (x - 4)$ rearranges to $y = 5 x - 20$ .
Fill in the two missing words in the method for estimating a derivative from points on a graph:
find the $______$ of the line segment joining two nearby points, taking one on each $______$ of the point in question.
The completed method is: find the slope of the line segment joining two nearby points, taking one on each side of the point in question.
The closer those two points are, the closer that secant slope comes to the slope of the tangent.
Points on a graph include $B (2, - 6)$ and $D (4, 0)$ . Use them to estimate $f^{'} (3)$ .
The estimate is $3$ .
The slope of the segment joining $B$ to $D$ is $\frac{0 - (- 6)}{4 - 2} = \frac{6}{2}$ , and $x = 3$ sits midway between the two points.
True or False?
When a table offers several pairs of values around a point, any pair straddling it gives an equally good estimate of the derivative.
False.
The pair that bounds the point most tightly gives the best estimate, because a shorter interval leaves the secant less room to drift away from the tangent.
Take the two values closest to the point, one on each side of it.
How do you estimate a derivative from a drawn tangent line?
Draw the tangent at the point, pick two points on the tangent line rather than on the curve, and work out the slope between them.
The accuracy depends entirely on how well the tangent was drawn, which is why the result is an estimate.
A table gives $f (2) = 4.5$ and $f (3) = 7.2$ among other values. Estimate $f^{'} (2.5)$ .
The estimate is $2.7$ .
The average rate of change between those two points is $\frac{7.2 - 4.5}{3 - 2} = 2.7$ , and $x = 2.5$ lies midway between them.
What must be true of a function before its derivative can be estimated from nearby points on its graph?
It must be continuous and differentiable across the interval containing those points.
Without that, the graph could do something between the two points that no secant slope would reveal.
Define a differentiable function.
A differentiable function is one whose derivative exists at every point of its domain.
A function with points where the derivative fails can still be made differentiable by restricting its domain so that those points are left out.
When does the derivative of $f$ exist at a point?
Exactly when the limit $\lim_{h \to 0} \frac{f (x + h) - f (x)}{h}$ exists at that point.
The derivative is defined by that limit, so wherever the limit fails there is no derivative to be had.
True or False?
A function that is continuous at a point must be differentiable there.
False.
Continuity is necessary for differentiability but nowhere near sufficient.
The graph of $f (x) = | x |$ is continuous at $x = 0$ and has no derivative there.
Name the four ways a continuous function can fail to be differentiable at a point.
The four are a corner, a cusp, a vertical tangent, and oscillation of the difference quotient near the point.
Each is a different way for the limit defining the derivative to fail there.
Fill in the two missing words distinguishing a corner from a cusp:
at a corner the two one-sided limits of the difference quotient are $______$ but unequal, while at a cusp one of them tends to $\infty$ and the other to $______$ instead.
The completed contrast is: at a corner the two one-sided limits of the difference quotient are finite but unequal, while at a cusp one of them tends to $\infty$ and the other to $- \infty$ instead.
At a vertical tangent, by contrast, both of them run to $\infty$ or both to $- \infty$ .
Show why $f (x) = | x |$ has no derivative at $x = 0$ .
The difference quotient there is $\frac{| h |}{h}$ , which equals $- 1$ for every $h < 0$ and $1$ for every $h > 0$ .
The two one-sided limits are $- 1$ and $1$ , so the limit defining $f^{'} (0)$ does not exist.
What does differentiability at a point tell you about continuity there?
If a function is differentiable at a point then it is continuous at that point.
In practice the implication is often used the other way round: a function that is not continuous somewhere cannot be differentiable there either.
A function $m$ is twice differentiable. What does that tell you about $m$ and $m^{'}$ ?
Both of them are continuous.
Because $m^{'}$ is itself differentiable it must be continuous, and $m$ being differentiable makes $m$ continuous in the same way.
Why is $f (x) = \frac{1}{x}$ not differentiable at $x = 0$ ?
Because $0$ is not in the domain of $f$ at all, so it cannot be in the domain of $f^{'}$ either.
A point at which the graph has a vertical asymptote can never be a point of differentiability.

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