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Chp 2 Review "R" help & "T" answers:
R3d) For this one, just find the limit by plugging in 6 for m. The "limit properties" they’re talking about are referring to the fact that you can also find the limit of the entire function by finding the limit of the individual pieces. One of the limit properties (that you don’t really need to know for the test) says that a limit of a function is the same of the sum of the limit of its pieces.
R3e) To find the average velocity from t = 5 to t = 5.1, you just need to find the slope between those two points:
d(5.1) - d(5) = -15.5 meters/sec
5.1 - 5
The average velocity from t = 5 to t sec will look the same as above, but with t plugged in instead of 5.1:
d(t) - d(5)
t - 5
When we find the limit of this expression as t approaches 5, it gives us the instantaneous velocity (instantaneous rate of change), because we're moving the two points t and t = 5 closer and closer together until they're in the same location.
To calculate this, plug in the equation for d(t) and find the value for d(5):
lim 35t - 5t^2 - (50)
t->5 t - 5
lim -5t^2 + 35t - 50
t->5 t - 5
Factor and simplify, then plug in 5 to find the limit:
lim -5(t^2 - 7t + 10)
t->5 t - 5
lim -5(t - 2)(t - 5)
t->5 t - 5
lim -5(t - 2)
t->5
= -5(5 - 2) = -15 meters/sec
The rock was going down at t = 5, because the velocity we just found is negative. An instantaneous velocity in this problem is the EXACT derivative.
T2a) -4 and -4
T2b) -4
T2c) Not continuous
T3a) No limit and No limit
T3b) No limit
T3c) Not continuous
T4a) 6 and 6
T4b) 6
T4c) Continuous
T5a) -2 and 3
T5b) No limit
T5c) Not continuous
T7a) [use your calculator to see the graph]
T7b) -7
T7c) [use your calculator to see the table, with TblStart=-3.05 and TblChange=0.01]
T7d) When delta = 0.02 (when x is between -3.02 and -2.98)
T8b) h(2) exists, the limit of h(x) as x approaches 2 exists, and h(2) = lim h(x) as x approaches 2.
T8c) h(3) = -23 and h(2) = 5. Because h(x) is continuous, and h(3) < 0 and h(2) > 0, according to the Intermediate Value Theorem there must be at least one zero in the interval 2 < x < 3.
Use your graphing calculator to find the zero "as accurately as possible": (2.194,0).
T9a) [use your calculator to see the table]
T9b) 0.901 cm/day
T9c) 0.01t + 0.7. Limit as t approaches 20 is 0.9 cm/day. This is the derivative!
T9c help) Start with the definition of derivative:
lim d(t) - d(20)
t—>20 t - 20
Plug in d(t), and find d(20) and plug it in:
lim 0.01t^2 + 0.5t - 14
t—>20 t - 20
Find a common factor to “factor the top”. Factor out 0.01 from each term:
lim 0.01(t^2 + 50t - 1400)
t—>20 t - 20
Then factor the quadratic equation like normal:
lim 0.01(t + 70)(t - 20)
t—>20 t - 20
You can cancel the (t - 20), then finally plug in t = 20 to find the exact derivative:
lim 0.01(t + 70) = 0.9 cm
t—>20
T9d) Seems to be speeding up, because it is covering more distance every 10 days.
T10) c(0) = 10 ft/s and p(0) = 10 ft/s. The limit of Calvin as t approaches infinity is 16, while the limit of Phoebe as t approaches infinity is "positive infinity" or in other words, "no limit". Phoebe is surprising! Who knew she was so infinitely fast?
T10 help) To find a limit at infinity, try plugging in a huge number, like t = 1,000,000. Think about what each answer is getting “closer to” as the number you plug in gets bigger.
Calvin: 16 - 6(2^-1,000,000) is the same as 16 - 6 / 2^1,000,000. This fraction at the end is REALLY small and is really close to 0. (Because 1/100 or 1/10,000 or 1/1,000,000 keeps becoming a smaller number, and gets closer to 0.) So, the Calvin equation can be thought about as:
16 - 6(0) = 16 ft/s
Phoebe: 10 + sqrt(1,000,000). This will just keep getting bigger as the input gets bigger. So, the answer is “infinity” or “no limit”.
T12) h(1) = 1 and h(2) = 8, so 7 is in between these two values. The intermediate value theorem says that there exists an x (in this case, x = cube root of 7) between x = 1 and x = 2 that will give you a value of 7.
2.7 Exploration Summary wkst.
8) The “difference quotient” is another name for the slope formula. It’s asking you to find the slope between two x-values: x = 1 and an unknown x.
f(x) - f(1)
x - 1
Graph this equation on your calculator, and follow the rest of the instructions given in the question.
Sudoku Puzzle (extra credit assignment):
C) Start by distributing to simplify the equation:
= 8x + 8h + 15 - 8x - 15
h
Then combine like terms and simplify:
= 8h
h
= 8
So, you’re actually just finding the limit of “8” as h approaches 0, so the answer is 8.
E) Try using synthetic division. Be sure to include a zero for every coefficient, even if there’s not a term included in the equation for it.
Set up:
1 | 1 0 0 0 0 -1
Next step:
1 | 1 0 0 0 0 -1
1 1 1 1 1
1 1 1 1 1 0
So, you’re finding the limit of x^4 + x^3 + x^2 + x + 1 as x approaches 0. Plug in x = 1 to find the limit:
1^4 + 1^3 + 1^2 + 1 + 1
= 1 + 1 + 1 + 1 + 1
= 5
F) Try the same strategy as part C.
Sec 2.6 IVT packet:
14) There's a little bit of guess and check here (when they don't give you an interval). A good place to start is with f(0) = 0. Now you just need to find another x-value that is greater than 0.4. It also needs to be on an interval that is continuous (it's OK if the entire function is not continuous).
I found this value by looking at the graph on my calculator: f(0.877) = 0.550.
Final conclusion: Because f(x) is continuous, and f(0) < 0.4 < f(0.877), then by IVT there exists an x = c in [0, 0.877] where f(c) = 0.4.
Sec 2.4:
67) "I am emailing you to clarify what the important concept in #67 you'd like us to master. I was able to do the problem without a calculator, and got the answers correct on the first try. What is the concept you'd like to make sure we mastered from this problem?"
I want everyone to make sure they get the answer correct. Even more than that, I want everyone to understand what that answer means when we're talking about the function being continuous.
Sec 2.3:
20c) Start by just writing down the equation they gave you: a(9) = [v(t) - v(9)] / (t - 9). Write it as a regular looking fraction, with a numerator and denominator. See how it looks like rate of change? It’s because acceleration “a” is the rate of change of velocity “v”.
You can find v(9) by plugging t = 9 into v(t). Then, plug this answer into your starting equation in place of v(9).
Remember when you learned to factor, and you learned the pattern that x^2 - 25 = (x + 5)(x - 5)? This factoring is called a “difference of two squares”, because it’s a difference (subtraction) of two squared things (x and 5). Here, try to factor t - 9 into “two squared things”. You’ll end up with a factored equation involving square roots:
t - 9 = (sqrt(t) + 3)(sqrt(t) - 3)
Now, try simplifying the top by factoring out a common factor. You should be left with a factor that cancels with the bottom. After canceling, you will be able to plug in t = 9 to find the limit as t approaches 9.
Email me for additional help and steps on this problem! :)
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