Differentiation Flashcards
In one word,
what is
differential calculus
about?
The
instant.
More specifically, the
rate of change at an instant.
What type of
line
will tell you the
average speed over time?
Secant line.
A
slope tells you the
ARC of a
vertical variable
with respect to a
horizontal variable.
ARC is
Δy
Δx
which is also the
slope of the secant line.
ex:
Δy
Δx
= y2 − y1
x2 − x1
= 4 − 0
1 − 0
= 4

What type of
line
will tell you
instantaneous speed?
Tangent line.
A
slope tells you the
ARC of a
vertical variable
with respect to a
horizontal variable.
The slope of the secant line will approach that of the tangent line as the distances for the secant line approach zero.

g(x) = √(x)
How can you
- *express** the
- *derivative** of √(x)?

A, C, and D.

What is a
- *reasonable** estimate of
- *g’(1)**?

A: −2
B: 1/4
C: 2
D: 0
E: −1/4
−2

Compare
f(−4) ___ f’(−1).
A: <
B: >

B: >
The slope of the tangent line would be less negative at f(−4).

The
- *tangent line** to the graph of
- *function f** at the point
- *(2, 3)** passes through the point
- *(7, 6)**.
What is
f’(2)?
3/5.
The derivative tells you the slope of the tangent.
Δy
Δx
= y2 − y1
x2 − x1
= 6 − 3
7 − 2
= 3/5
What is the
- *formal definition** of the
- *derivative** of a function?
f’(x) = _____?
limΔx→0 f(x0 + Δx) − f(x0)
Δx
This is basically the slope formula with a limit tacked on:
Δy
Δx
= y2 − y1
x2 − x1
= limΔx→0 f(x0 + Δx__) − f(x0)
(x0 + Δx) − x0
= limΔx→0 f(x0 + Δx) − f(x0)
Δx

What is the
- *alternate definition** of the
- *derivative** of a function?
f’(x) = _____?
limx→a f(x) − f(a)
x − a
This is basically the slope formula with a limit tacked on:
Δy
Δx
= y2 − y1
x2 − x1
= limx→a f(x) − f(a)
x − a

How do you write the
equation of a
tangent line using the
formal definition of a
limit of the function below?
f(x) = x2 at x = 0.2

-
Find the derivative
f(x) = x2
limΔx→0 f(x + Δx) − f(x)
(x + Δx) − x
f’(x) = limΔx→0 (x + Δx)2 − x2
(x + Δx) − x
= limΔx→0 x2 + 2x•Δx + Δx2 − x2
Δx
= limΔx→0 2x•Δx + Δx2
Δx
= limΔx→0 2x + Δx
f’(x) = 2x -
Find the slope of the tangent
f’(x) = 2x
= 2(0.2)
f’(0.2) = 0.4 -
Find the right point on the function
f(x) = x2
f(0.2) = (0.2)2
f(0.2) = 0.04 -
Write the equation of the tangent line
y − y0 = m(x − x0)
y − 0.04 = 0.4(x − 0.2)

Given:
- f(−1) = 2
- f(0) = 0
- f(1) = 1
- f(8) = 5
what is the
best estimate of f’(−1/2) we can make
given these values?
−2
Δy
Δx
= y2 − y1
x2 − x1
= 2 − 0
−1 − 0
= −2
Given:
- f(−1) = 2
- f(0) = 0
- f(1) = 1
- f(8) = 5
what is the
best estimate of f’(8) we can make
given these values?
4/7
Δy
Δx
= y2 − y1
x2 − x1
= 5 − 1
8 − 1
= 4
7
Graphically,
what are
all x-values for which this
function is
not differentiable?
Dashed lines represent asymptotes.

−5, −4, 0

Graphically,
what are
all x-values for which this
function is
not differentiable?
There’s a
vertical tangent at x = 1.

1, 5

Visually, what
features of a
graph will indicate that
the function is
not differentiable at an x-value?
-
Vertical tangent
* (the slope of the tangent line is undefined)* -
Discontinuity
* (there’s no limit at that location)* -
“Sharp” turn
* (the one-sided limits aren’t equivalent, so there’s no limit at that location)*
How do you know whether function
- *g(x)** is
- *continuous** at
- *x = a**?
g(x) is continuous at x = a if
- *g(a)** and
- *both one-sided limits** are
- *equal**.
- (Which is to say that the limx→a g(x) exists and is equal to g(a).)*
- ex:*
g(x) = { x2 + 2x , x < 1
{ 4x − 1 , x > 1
g(1) = (1)2 + 2(1)
= 3 = limx→1− g(x)
limx→1+ g(x) = 4(1) − 1
= 3
g(1) = limx→1− g(x) = limx→1+ g(x)
So g(x) is continuous at x = 1.
How do you know whether function
- *g(x)** is
- *differentiable** at
- *x = a**?
g(x) is differentiable at x = a
if it is
continuous at x = a and
limx→a f(x) − f(a)
x − a
exists.
(Which is to say that you can find the slope of a tangent line that intersects g(x) at x = a)
(Once you know it’s continuous, you’re looking for a
vertical tangent or a
“sharp” turn)
ex:
g(x) = { x2 + 2x , x < 1
{ 4x − 1 , x > 1
g(1) = (1)2 + 2(1)
= 3 = limx→1− g(x)
limx→1+ g(x) = 4(1) − 1
= 3
g(1) = limx→1− g(x) = limx→1+ g(x)
So g(x) is continuous at x = 1.
limx→a f(x) − f(a)
x − a
limx→1− x2 + 2x − (3)
x − 1
= limx→1− (x − 1)(x + 3)
x − 1
= limx→1− x + 3
= 4
limx→1+ 4x − 1 − (3)
x − 1
= limx→1+ 4x − 4
x − 1
= limx→1+ 4(x − 1__)
x − 1
= 4
Because the one-sided limits are approaching the same value, the limit exists, and g(x) is differentiable at x = 1.
Differentiability
implies _____.
Continuity
Proof:
-
Differentiability:
limx→c f(x) − f(c) = f’(c)
x − c -
Continuity:
limx→c (f(x) = f(c))
Assume f is differentiable at x = c
limx→c (f(x) − f(c))
= limx→c (x − c) • f(x) − f(c)
x − c
= limx→c (x − c ) • limx→c f(x) − f(c)
x − c
= 0 • limx→c f(x) − f(c)
x − c
= 0 • f’(c)
limx→c (f(x) − f(c)) = 0
limx→c f(x) − limx→c f(c) = 0
limx→c f(x) − f(c) = 0
limx→c f(x) = f(c)
(this is the definition of continuity)
Constant Rule:

0.
The derivative of
any constant is 0.
Algebraically:
f(x) = 1
f(x + Δx) = 1
g’(x) = limΔx→0 f(x + Δx) − f(x)
Δx
f’(x) = limΔx→0 1 − 1
Δx
= limΔx→0 0
Δx
= 0
Graphically:
The derivative measures a function’s
instantaneous rate of change at a particular x-value.
Where f(x) is a constant, there is
no change from one x-value to the next, so the derivative is 0.

Sum/Difference Rules:
f(x) = g(x) + j(x)
f’(x) = ______
g’(x) + j’(x)
f(x) = g(x) + j(x) ⇒ f’(x) = g’(x) + j’(x)
Algebraically:
f(x) = g(x) + j(x)
f’(x) = limΔx→0 g(x + Δx) + j(x + Δx) − (g(x) + j(x))
Δx
= limΔx→0 g(x + Δx) − g(x) + j(x + Δx) − j(x)
Δx
= limΔx→0 g(x + Δx) − g(x) + j(x + Δx) − j(x)
Δx Δx
= limΔx→0 g(x + Δx) − g(x) + limΔx→0 j(x + Δx) − j(x)
Δx Δx
= g’(x) + j’(x)
The difference rule is identical.
Constant Multiple Rule:
- *d [k•f(x)]** = _____?
- *dx**
k• d [f(x)]
dx
Also:
f(x) = k•g(x) ⇒ f’(x) = k•g’(x)
Algebraically:
h’(x) = limΔx→0 f(x + Δx) − f(x)
Δx
f(x) = k•g(x)
f’(x) = limΔx→0 k•g(x + Δx) − k•g(x)
Δx
= limΔx→0 k•g(x + Δx) − k•g(x)
Δx
= limΔx→0 k • g(x + Δx) − g(x)
Δx
= k•limΔx→0 g(x + Δx) − g(x)
Δx
= k•g’(x)
Constant Rule:
f(x) = 2x3 − √x + 1/x + 2
f’(x) = _____
6x2 − ½x−1/2 − x−2
*f(x) = x<sup>n</sup> f'(x) = nx<sup>n−1</sup>*
f(x) = 2x3 − √x + 1/x + 2
= 2x3 − x1/2 + x−1 + 2
f’(x) = d/dx(2x3) − d/dx(x1/2) + d/dx(x−1) + d/dx(2)
= (3)2x3−1 − (½)x1/2−1 + (−1)x−1−2 + 0
= 6x2 − ½x−1/2 − x−2
Power Rule:
f(x) = xn, n ≠ 0
f’(x) = _____
f’(x) = nxn−1

Proof:
(for positive integers)
f(x) = 2x3 − √x + 1/x + 2
f’(x) = _____
6x2 − ½x−1/2 − x−2
*f(x) = x<sup>n</sup> f'(x) = nx<sup>n−1</sup>*
f(x) = 2x3 − √x + 1/x + 2
= 2x3 − x1/2 + x−1 + 2
f’(x) = d/dx(2x3) − d/dx(x1/2) + d/dx(x−1) + d/dx(2)
= (3)2x3−1 − (½)x1/2−1 + (−1)x−1−2 + 0
= 6x2 − ½x−1/2 − x−2
How would you write the
- *equation** of a
- *line**
- *tangent** to
f(x) = x3 − 6x2 + x − 5
at
x = 1?
-
Find the point the tangent touches
(this is just f(1))
f(x) = x3 − 6x2 + x − 5
f(1) = (1)3 − 6(1) + (1) − 5 = −9
(1, −9) -
Find the slope of the tangent at that point
(this is just the derivative)
f’(x) = 3x2 − 12x2 + 1
f’(1) = 3(1)2 − 12(1)2 + 1
= −8 -
Write the equation of the line
y − y0 = m(x − x0)
y + 9 = −8(x − 1) (for point/slope form)
y = −8x + 8 − 9
y = −8x − 1 (for slope/intercept form)

f(x) = sin(x)
f’(x) = _____

cos(x)

Green: slope of 1
Orange: slope of 0
Red: slope of −1
f(x) = cos(x)
f’(x) = _____

−sin(x)

Green: slope of 1
Orange: slope of 0
Red: slope of −1
f(x) = ex
f’(x) = _____

ex
- This is one of the “magical” things about e.*
- In fact, e can be defined this way.*

f(x) = ln(x)
f’(x) = _____

1
x

Product Rule:
p(x) = [f1(x) • f2(x)]
p’(x) = _____
f1‘(x) • f2(x) + f1(x) • f2‘(x)
p(x) = [ln(x) • cos(x)]
p’(x) = d/dx(ln(x)) • cos(x) + ln(x) • d/dx(cos(x))
= 1/x • cos(x) + ln(x) • (−sin(x))
= cos(x) − ln(x) • sin(x)
x
Quotient Rule:
q(x) = [n(x)]
[d(x)]
q’(x) = _____
n’(x) • d(x) − n(x) • d’(x)
( d(x) )2
q(x) = [sin(x)]
[x2]
q’(x) = d/dx(sin(x)) • x2 + sin(x) • d/dx(x2)
(x2)2
= x2 • cos(x) − sin(x) • 2x
x4
= x (x•cos(x) − 2•sin(x))
x4
= x•cos(x) − 2•sin(x)
x3
f(x) = tan(x)
f’(x) = _____
1 = sec2(x)
cos2(x)
f(x) = tan(x)
= sin(x)
cos(x)
f’(x) = d [sin(x)]
dx [cos(x)]
= cos(x) • cos(x) + sin(x) • sin(x)
cos2(x)
= cos2(x) + sin2(x)
cos2(x)
= 1
cos2(x)
= sec2(x)
f(x) = cot(x)
f’(x) = _____
− 1 = −csc2(x)
sin2(x)
f(x) = cot(x)
= cos(x)
sin(x)
f’(x) = d [cos(x)]
dx [sin(x)]
= −sin(x) • sin(x) − cos(x) • cos(x)
sin2(x)
= −sin2(x) − cos2(x)
sin2(x)
= − 1
sin2(x)
= −csc2(x)
f(x) = sec(x)
f’(x) = _____
tan(x) • sec(x)
f(x) = sec(x)
= 1
cos(x)
= 0 • cos(x) + sin(x) • 1
cos2(x)
= sin(x)
cos2(x)
= sin(x) • 1
cos(x) cos(x)
= tan(x) • sec(x)
f(x) = csc(x)
f’(x) = _____
−cot(x) • csc(x)
f(x) = csc(x)
= 1
sin(x)
= 0 • sin(x) − cos(x) • 1
sin2(x)
= −cos(x)
sin2(x)
= −cos(x) • 1
sin(x) sin(x)
= −cot(x) • csc(x)
Chain Rule:
f(x) = o(i(x))
f’(x) = _____
o’( i(x)) • i’(x)
ex:
- h(x) = (5 − 6x)5 = o(i(x))
- i(x) = 5 − 6x
i’(x) = −6 - o(x) = x5
o’(x) = 5x4
- i(x) = 5 − 6x
- h’(x) = o’(i(x)) • i’(x)
- = 5 ( 5 − 6x)4 • −6
- −30( 5 − 6x)4
Is
f(x) = cos2(x)
a
composite function?
Yes,
so the
chain rule applies.
A function is composite if you can write it as
f(g(x)).
In other words, it is a
function within a function, or a
function of a function.
cos2(x) = (cos(x))2 = o(i(x))
i(x) = cos(x)
o(x) = x2
f(x) = bx,
where b is any positive number
d [bx] = _____
dx
bx • ln (b)
d/dx [ex] = ex
b = eln (b)
(remember: logb a = c ⇔ ba = c)
(ln (b) is the number that you have to raise e to in order to get b, so if you raise e to that number, you get b)
d/dx [bx]
= d/dx [(eln (b))x] (substitution)
= eln (b) • x • ln(b) (chain rule)
= (ln (b)) • (eln (b))x
= bx • ln (b)
f(x) = logb a,
where b is any positive number
d/dx [logb a] = _____
1
(ln b) • a
d/dx [ln x] = 1/x
- *d [y]** = _____?
- *dx**
dy
dx
These are simply equivalent statements.
- *d [y2]** = _____?
- *dx**
2y • dy
dx
Just the chain rule:
c(x) = o(i(x))
c’(x) = o’(i(x) • i’(x)
- First, we assume that y does change with respect to x — it’s not a constant.
- Then you can set up y as a function of x:
y(x)
= d [y2]
dx
= d(y2) • dy
dy dx
↑ This part is just the chain rule:
The
derivative of *something* squared
with respect to that *something*
times the
derivative of that *something*
with respect to x.
= 2y • dy
dx
- *d [xy]** = _____?
- *dx**
y + x • dy
dx
Apply the product rule:
p(x) = f1(x) • f2(x)
p’(x) = f1‘(x) • f2(x) + f1(x) • f2‘(x)
- *d [xy]**
- *dx**
= dx • y + x • dy
dx dx
= y + x • dy
dx
Given
x2 + y2 = 1,
how do you obtain
- *dy**?
- *dx**
Implicit differentiation:
Treat the
variable you’re looking for as a
function of the other and
solve.

x2 + y2 = 1
d (x2 + y2) = d (1)
dx dx
d (x2) + d (y2) = 0
dx dx
2x + 2y • dy = 0
dx
2y • dy = −2x
dx
dy = − x
dx y
d sin−1 (x) = _____
dx
1
√(1 − x2)
Equivalent statements:
y = sin−1 (x) ⇔ sin(y) = x
x = sin(y)
Derivative of Both Sides w/ respect to x:
dx/dx = d/dx [sin(y)]
1 = d/dy [sin(y)] • dy/dx
1 = cos(y) • dy/dx
_dy_ = _1_ dx cos(y)
Pythagorean trig identity:
dy = 1
dx √(1 − sin2(y))
Substitution:
dy = 1
dx √(1 − x2)
d cos−1 (x) = _____
dx
− 1
√(1 − x2)
Equivalent statements:
y = cos−1 (x) ⇔ cos(y) = x
x = cos(y)
Derivative of Both Sides w/ respect to x:
dx/dx = d/dx [cos(y)]
1 = d/dy [cos(y)] • dy/dx
1 = −sin(y) • dy/dx
dy = 1
dx − sin(y)
Pythagorean trig identity:
dy = − 1
dx √(1 − cos2(y))
Substitution:
dy = − 1
dx √(1 − x2)
d tan−1 (x) = _____
dx
1
1 + x2
Equivalent statements:
y = tan−1 (x) ⇔ tan(y) = x
x = tan(y)
Derivative of Both Sides w/ respect to x:
dx/dx = d/dx [tan(y)]
1 = d/dy [tan(y)] • dy/dx
1 = 1 • dy/dx
cos2(y)
dy = cos2(y)
dx
Pythagorean trig identity:
= cos2(y)
cos2(y) + sin2(y)
Multiply numerator and denominator by same amount:
= cos2(y) • 1 / cos2(y)
cos2(y) + sin2(y) 1 / cos2(y)
= 1
1 + ( sin(y) / cos(y) )2
Def of tangent:
= 1
1 + tan2(y)
Substitution:
= 1
1 + tan2(y)
Functions
f and g are
inverses.
In
terms of its inverse,
f’(x) = _____.
1
g’(f(x))
The
derivative of a function is
equal to the
reciprocal of the
derivative of its
inverse.
- Definitions
f(x) = g−1(x) ⇔ g(x) = f−1(x)
f(g(x)) = x = g(f(x)) - Now, the work . . .
- g(f(x)) = x
-
d [g(f(x))] = dx
dx dx - g’(f(x)) • f’(x) = 1
- f’(x) = 1
g’(f(x))

Given the
table below,
and that
functions g and h are
inverses,
how could you
determine
g’(0)?

g’(0) = 1
h’(g(0))

g’(x) = 1
h’(g(x))
= 1
h’(g(0))
= 1
h’(−2)
= 1
−1
= −1
What’s the best way to
evaluate
- *d [(x + 5)(x − 3)]**?
- *dx**
Expand,
then the
product rule.
Expressions can be rewritten to make differentiation easier!
What’s the
- *key** to
- *evaluating**
- *d [(x2 sin(x))3]**?
- *dx**
Label, label, label.
When applying multiple rules, label them all.
-
Rewriting:
(x2 sin(x))3
= x6 sin3(x)
= f(x) • g(h(x)) -
Labeling:
f(x) = x6
g(x) = x3
h(x) = sin(x) -
Differentiating:
f’(x) = 6x5
g’(x) = 3x2
h’(x) = cos(x) -
d [x6 sin3(x)]
dx -
Product Rule:
= d [f(x)] • g(h(x)) + f(x) • d [g(h(x))]
dx dx -
Chain Rule:
= f’(x) • g(h(x)) + f(x) • g’(h(x)) • h’(x) -
Substitution:
= 6x5 • sin3(x) + x6 • 3 sin2(x) • cos(x) - = 6x5 sin3(x) + 3x6 sin2(x) cos(x)
What’s the
- *key** to
- *evaluating**
- *d [sin (ln (x2))]**?
- *dx**
Label, label, label.
When applying multiple rules, label them all.
- sin(ln(x2)) = f (g (h(x)))
- f(x) = sin(x)
g(x) = ln(x)
h(x) = x2 - f’(x) = cos(x)
g’(x) = 1/x
h’(x) = 2x -
d [sin(ln(x2))]
dx - = f’ (g (h(x))) • g’ (h(x)) • h’(x)
- = cos (ln (x2)) • 1/(x2) • 2x
- = 2 cos (ln (x2))
x
f(x) = x3 + 2x2
f’(x) = 3x2 + 4x
f’‘(x) = _____
6x + 4
This is the
second derivative, or the
derivative of the first derivative
Could also be written as:
d2y [x3 + 2x2]
dx2
How would you
evaluate
limh→0 5 log (2 + h) − 5 log (2)
h
Use the derivative
limh→0 5 log (2 + h) − 5 log (2)
h
= 5 limh→0 log (2 + h) − log (2)
h
- The above, by definition, is f’(2)*
- So…*
- f(x) = log(x)*
f’(x) = 1
ln(10) x
f’(2) = 1
ln(10) 2
Substitution:
5 • f’(2) = 5 limh→0 log (2 + h) − log (2)
h
= 5
2 • ln(10)
What is a
related rates problem?
Applied problems where
you find the
rate at which
one quantity is changing by
relating it to
other quantities whose
rates are known.
e.g.:
The radius r(t) of a circle is increasing at a rate of 3 centimeters per second. At a certain instant t0, the radius is 8 centimeters.
What is the rate of change of the area A(t) of the circle at that instant?
What are the
- *keys** to a
- *related rates problem**?
- Draw a diagram
-
Label, label, label
Label what you know and what you’re trying to figure out -
Divide and conquer
Use information you know to figure out information you need — in several steps if necessary.
Given
√(2) = 4,
how might you
approximate
√(4.36)?

Local linearity

You can use the line tangent to at a known point to approximate the value of a value that’s close, but unknown.
In words,
What can
L’Hôpital’s rule
do?
Use
- *derivatives** to find
- *limits** with
- *undetermined forms**
This is in contrast to what we normally do, which is to use limits to find derivatives (in fact, the definition of derivatives is based on limits).
Mathematically,
what is
L’Hôpital’s rule?
Zero Case:
- If:
limx→c f(x) = 0
&
limx→c g(x) = 0
&
limx→c f’(x) = L
g’(x) - Then:
limx→c f(x) = L
g(x)
Infinity Case:
- If:
limx→c f(x) = ±∞
&
limx→c g(x) = ±∞
&
limx→c f’(x) = L
g’(x) - Then:
limx→c f(x) = L
g(x)
How would you
evalulate
- *limx→0 sin(x)**?
- *2x**

L’Hôpital’s rule

Direct substitution yields:
limx→0 sin(x) = 0
2x 0
0/0 is undetermined form.
The first two conditions of the rule are met, so we should try to take the derivatives to see if the third condition is also met.
= limx→0 cos(x) = 1
2 2
Therefore . . .
limx→0 sin(x) = 1
2x 2
How would you
evalulate
- *limx→∞ 4x2 − 5x**?
- *1 − 3x2**
L’Hôpital’s rule
(is one way, although factoring and canceling would also work)
Direct substitution yields:
limx→∞ 4x2 − 5x = ∞
1 − 3x2 −∞
This is undetermined form.
The first two conditions of the rule are met, so we should try to take the derivatives to see if the third condition is also met.
= limx→∞ 8x − 5 = ∞
−6x −∞
This is still undetermined, so we should try L’Hôpital’s rule again:
limx→∞ 8 = −4
−6 3
Mathematically,
describe the
mean value theorem?
Given that function f is:
- differentiable over the
- open interval (a*, b)** and
- continuous over the
- closed interval [a*, b]**,
There is at least
- *one number c** on the
- *interval** from a to b such that
f’(c) = f(b) − f(a)
b − a

Graphically,
describe the
mean value theorem?
An
arc between
two endpoints has a
point at which the
tangent to the arc is
parallel to the
secant through its endpoints

Does the
- *mean value theorem**
- *apply** to this function?

No.
For the MVT to apply, the
function must be differentiable
over (a, b).
The function has only two possible tangent lines, neither of which is parallel to the secant between x = a and x = b.

Does the
- *mean value theorem**
- *apply** to this function?

No.
For the MVT to apply, the
function must be continuous
over [a, b].
All possible tangent lines are necessarily decreasing, while the secant line is increasing.

Does the
mean value theorem
apply to this function
over the interval
[−6, −2]?

There is a vertical tangent at x = −5
No,
because of the
vertical tangent.
The function
must be differentiable
over (−6, −2) for the MVT to apply.

Does the
mean value theorem
apply to this function
over the interval
[−4, −1]?

There is a vertical tangent at x = −5
Yes.
The function is continuous over [−4, −1] and differentiable over (−4, −1).

Does the
mean value theorem
apply to this function
over the interval
[−1, 2]?

There is a vertical tangent at x = −5
No.
The function
must be continuous
over (−1, 2) for the MVT to apply.

Does the
mean value theorem
apply to this function
over the interval
[0, 4]?

There is a vertical tangent at x = −5
Yes.
The function is continuous over [0, 4] and differentiable over (0, 4).

Does the
mean value theorem
apply to this function
over the interval
[0, 5]?

There is a vertical tangent at x = −5
No,
because of the
“sharp” turn.
The function
must be differentiable
over (0, 5) for the MVT to apply.

Mathematically,
describe the
extreme value theorem?
Given that function f is:
- *continuous** over the
- closed interval [a*, b]**,
There are
values c and d
in the interval [a, b] such that
f(c) < f(x) < f(d).

Graphically,
describe the
extreme value theorem?
Where function f is
continuous over [a, b],
there is an
absolute maximum value of f over the interval
and an
absolute minimum value of f over the interval.

Visualize why an
interval must be continuous
for the
extreme value theorem to apply.

Visualize why an
interval must be closed
for the
extreme value theorem to apply.

What is a
critical point?
An x-value:
- within the domain of function f (graphed below)
- where
-
f’(x) = 0
or - f’(x) is undefined
-
f’(x) = 0
Green is local max — f’(x) is undefined
Red is local min — f’(x) is undefined
Orange is global max — f’(x) is zero

How do you find an
increasing or decreasing interval
of a function?
Differentiate
and evaluate
around critical points.
f’(x) will be
positive where f(x) is increasing
and
negative where f(x) is decreasing.
How do you find
global/absolute extrema?
(aka the first derivative test)
Same as local/relative extrema,
except you must
check the edges
in both directionsbecause
endpoint can be global/absolute extrema.
- Differentiate
-
Find
-
critical points
and - where f(x) is undefined
-
critical points
- Analyze intervals using first derivative
-
Find extremum points, which will be where
-
f(x) is defined
and - f’(x) changes signs
-
f(x) is defined
- Check edges
How do you find
local/relative extrema?
(aka the first derivative test)
- Differentiate
-
Find
-
critical points
and - where f(x) is undefined
-
critical points
- Analyze intervals using first derivative
-
Find extremum points, which will be where
-
f(x) is defined
and - f’(x) changes signs
-
f(x) is defined