Unit 3 Quiz Flashcards
The slope of one function is what(of the derivative of that function)
The y value of the derivative of that function
If you identify the horizontal tangent lines of f you can find what for f’
You can find the zeroes of f’
to don’t relative min value(aka why value) what do u do with the x value your already found
plug back into the OG equation
2nd derivative test mathematically
first use 1st derivative to find critico al numbers
then find second derivative
plug in those critical numbers into second derivative
if you get a positive number that means it is concave up
if you get a negative number that means it is concave down
if f is constant what is the first derivative of f
0(slope of f)
if f is concave up what is f’
increasing
if f is concave down what is f”
negative
When is f” 0on a behavior table
(The options are pos, neg, 0)
Think of f graph in relation to
It is 0 when f is linear
There is no value of X in the open interval -1 to 3 at which a person of X is equal to blank explain why this does not violate the mean value theorem
Consider cusps in holes in the graph
If it’s not differentiable X equals 02 to a sharp corner does the mean value theorem cannot be applied on the interval -1 to 3
When asked if the graph of f is decreasing and concave down do what
Make BOTH sign charts then align the numbers!
When using calculator to graph both h’ and h’’ do what
Draw your own graph and overlap the two so you can see the visuals at the same time, make sure to identify critical points too
How to find POI on f(for f)
In the middle of the change fo slopes - that point where there is a vertical tan line
How to find the POI of f (using f’)
Min and max on f’ graph
How to find POI for f (using f’’)
Points on the x-axis, the zeroes
How to find relative extrema-min/ max for f (using f’)
It is the critical points
Extreme value theorem
Extreme aka absolute
A f(x) THAT is continuous on closed interval must have an absolute max or min
How to write extreme value theorem FRQ
If f is continuous on [-1,3] then there exists a number c such that f(c)greate than or equal to( for an absolute max) or less than or equal to (for an absolute min) for all #’s x in [a,b]
MVT
If f is continuous on [a,b] and differentiable on (a,b) than there exists at least 1 #c such that f’c = f(b)-f(a)/b-a
Idea of MVT
The instantaneous rate of change at one point should be equal to the average rate of change
1st derivative test
A critical point of f(x) is a relative min if f’(x) changes from negative to positive and for relative max vice versa
Where are the values for absolute maximum and min
Endpoints or critical interior points
What is first derivative test used for
Relative min and max
What test do you sue for relative mina and max
First derivative
What test do you use for extrema, absolute max/min?
Candidates test
Candidates test
Still use f’ for candidates test
Identify critical points on the graph as endpoints or relative min and max aka f’ zeroes
What si the order of the derivative
Increasing pos
Increasing/ deceasing - positive/negative - Ccu/ccd
The graph of f is concave up when it’s derivative
Is increasing
Points of inflection are
Extreme rates of change, max/ min value of a derivative
2nd derivative test rules criteria(can on,y be used when …)
The 2nd D can only be used when
F has a critical point at x=a and f’(a)=0
The 2nd D test does NOT work When (think ab graph of f)
f”(a)>0 or vice versa
Or if the slope(f’) is undefined
When do we use 2nd D test most often amd why
On problems with tables because we can’t compare values of f’ to the left and right of the critical point
following functions of x is guaranteed by extreme value thrm to have an absolute max on the interval consider
consider if those functions can be undefined in denominator, if so they do not work
fewest possible number of values of c in the MVT such that f’c = 6
try every average rate of change for all values since instantaneous rate(f’c) of change equals average rate of change such that you get 6
when dealing with mean value theorem MCQ answer must have
instantaneous rate of change and average rate of chabge
derivative of e^x
e^x
if f(x) is 1/x^2 and you are asked to find the min/max valued for f on [0,3] what do you do FRQ
realize that f has a hole at 0 which is between 0 and 3 therefore the EVT can’t be applied you’d say: since f(x) is rational func and continuous everywhere except x=0 the EVT cant be applied
where f’(x)=0 and f’(x)= DNE
what are these points called
critical numvers
for candidates test what do you do in your chart
put in values of critical and endpoints then plug those into the ig function and the highest y value is the absolute max and the lowest is the absolute min
derivative of lnx
1/x
what is the domain for ln x
x must be greater than 0
tangent line is what rate of change
instantaneous
secant lien is what rate of change
average
how do you prove f(x) is differentiable in MVT
find f’(x) see if it’s differentiable(aka polynomial not rational)
To find all values of c in MVT
do what
1) write f(x) is continuous on closed interval
2) find f’(x) and if so write that it’s differentiable on open interval
3) do f(b)-f(a)/b-a and set it equal to f’(x)
4) solve for x
5) x is the same thing as c as long as it fits in the interval
cos x = 0
what is x
x is radians
pi/2, 3pi/2
cosx-sinx cna be simplified to
cos x = sin x -> cosx/cosx = sinx/cosx -> 1=tan x
