review 1 part 2 limits part 2. limits recognizing that when you have a limit, you don’t always...
TRANSCRIPT
Review 1 Part 2
Limits part 2
Limits
• Recognizing that when you have a limit, you don’t always have to evaluate it. Chances are, when you evaluate it, you will generate a hole in the graph – 0/0.
What is this asking you to do?
lim ( )
h
hh
0
1 17
• If you plug in 0 for h, you get 0/0 when done evaluating it. You don’t want to foil out 1 +h seven times….so you can do 2 things…..apply L’hospitals rule or recognize that it is the definition of a derivative at a point. Recognizing it the fastest method.
• So the equation is and you are evaluating it at 1. So the question is asking for the tangent slope of the curve at x = 1. Take the derivative and plug in 1. You get 7.
x 7
Another tangent slope
• Given
• A- what is the function?• B – what x-coordinate are you evaluating it
at?
lim( )
h
e eh
h
0
2 2 4
• The function must be e^? So what are you raising e to? The limit has the h + 2 being squared. And there is a 2 with the h, not a 4.
• 2 squared it 4, so is really e 4 e 22
• So the equation must be
• The derivative would be• Evaluating it at x = 2 gives you
e x2
22
xe x
4 4e
Definition of the slope of a curve at a point
lim( ) ( )
h
f a h f ah
0
Alternate Definition
• The alternate definition is asking the same thing. What is the tangent slope, but it uses a form of a secant slope, keeping the equation in the numerator and an x in the denominator.
• Given
• This can be evaluated by using L’hospitals rule or by recognizing that the function is the square root of x and that you are looking for the slope at x = 2.
limx
xx
2
22
• So the derivative would be
• Evaluating at x = 2 you get
12 x
12 2
Alternate Derivative
lim( ) ( )
x a
f x f ax a
• The previous examples show up in multiple choice questions! They are designed to take minimal time, provided you recognize what they are asking. They can also show up in questions that have I, II, and III in the question, and you are asked to determine which is correct.
Free response question using a limit
• A car is traveling on a straight road. For 0 < t< 24 seconds, the car’s velocity v(t), in meters per second, is modeled by the piecewise-linear function defined by the graph above.
• B) For each of v’(4) and v’(20), find the value or explain why it does not exist. Indicate units of measure.
• This question is asking you to find a secant slope and explain it using limits!
• The graph has a corner at t = 4, so v’(4) does not exist. You explain this using limits. i.e. find the left and right side limits, showing that as slopes, they do not match!
• So one slope is 5 and the other slope is 0. The write up would be,
• The limits are not the same, therefore, v’(4) DNE.• You could also explain that the slopes on the left
and right side are different – thus v’(4) DNE.
lim( ) ( )
t
v t v
t
4
4
45
lim( ) ( )
t
v t v
t
4
4
40
• V’(20) is on the line – so you only have to show that it is a secant slope.
v m( ) / sec20 20 016 24
52
2
• D) Find the average rate of change (that means secant slope) of v over the interval 8< t< 20. Does the Mean Value Theorem guarantee a value c, for 8 < c< 20, such that v’(c) is equal to this average rate of change? Why or why not?
• The question is asking 2 things: Find a secant slope of v and then explain if the mean value theorem holds over the interval.
• The mean value theorem can apply if: The function is continuous over the given interval and that the function has a derivative over the interval.
• Step 1 – find the average rate of change:
• You need to show the first part and any of the three parts that follow. You do not have to simplify. But you need to include units.
avera tef b f a
b a
( ) ( )
v v m( ) ( ) / sec20 820 8
10 2020 8
1012
56
2
• Does the mean value theorem apply. No. The mean value theorem does not apply to v on the given interval since v is not differentiable at t = 16 (corner).