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Calculus Review.

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Presentation on theme: "Calculus Review."β€” Presentation transcript:

1 Calculus Review

2 How do I know where f is increasing?
It is where f prime is positive. Find the derivative and set equal to zero. Use test points to find where f prime is positive or negative.

3 How do I know where f has inflection points?
It is where f double prime equals zero or is undefined and the sign changes. The f prime function changes from increasing to decreasing or vise-versa. It is where f prime has maximums or minimums.

4 How do I know if the particle is moving to the left?
It is where f prime is negative. Find where f prime equals zero. Then check test points on f prime.

5 How do I know if the particle is speeding up or slowing down?
Find v(t) and a(t): If they have the same sign the particle is speeding up. If they have different signs the particle is slowing down.

6 What is speed? It is the absolute value of velocity.

7 What do I do if the problem says find the particular solution y = f(x)?
This is asking you to find the original function that represents f. You are probably doing a separable variable problem.

8 What does 𝑑𝑦 𝑑π‘₯ mean? This is asking for the first derivative.
It could also be written 𝑑 𝑑π‘₯

9 What does 𝑑 2 𝑦 𝑑π‘₯Β² mean? This is asking for the second derivative.
It could also be written 𝑑 2 𝑑 π‘₯ 2 .

10 What is the limit definition of a derivative?
Or 𝑓 β€² π‘Ž = lim π‘₯β†’π‘Ž 𝑓 π‘₯ βˆ’π‘“(π‘Ž) π‘₯βˆ’π‘Ž

11 What are the limit rules?
Is it a hidden derivative? If it’s approaching infinity and it’s a polynomial over a polynomial then use the horizontal asymptote rules. Can you factor to simplify and just plug in the numbers. L’Hopital’s rule: do f(x) and g(x) both approach 0 or ±∞, then lim π‘₯β†’π‘Ž 𝑓(π‘₯) 𝑔(π‘₯) = lim π‘₯β†’π‘Ž 𝑓′(π‘₯) 𝑔′(π‘₯) .

12 How do I prove a function is continuous?
Show the left hand limit and the right hand limit are equal and are equal to f(x).

13 Day One Practice

14 Differentiate: arctan 2x
2 1+4π‘₯Β²

15 Differentiate: 𝑒 π‘₯ 𝑒 βˆ’π‘₯

16 Differentiate: 2 3π‘₯ 3Γ— 2 3π‘₯ Γ—ln⁑(2)

17 𝑦= 1 π‘₯ , Find 𝑑𝑦 𝑑π‘₯ βˆ’2π‘₯ π‘₯ βˆ’2

18 𝑓 π‘₯ = π‘₯βˆ’1 π‘₯ , find 𝑓 β€² π‘₯ . π‘₯ π‘₯ 2 βˆ’8π‘₯+1

19 sin 2π‘₯ + cos 3π‘₯ 𝑑π‘₯= βˆ’cos⁑(2π‘₯) 2 + sin⁑(3π‘₯) 3 + C

20 π‘₯ π‘₯ 2 βˆ’6 dx= 1 2 lnβƒ’ π‘₯ 2 βˆ’6βƒ’+𝐢

21 𝑓 β€²β€² π‘₯ = π‘₯ 2 π‘₯βˆ’4 π‘₯βˆ’8 . Find the x-coordinate(s) for points of inflection on f.
x=4 and x=8

22 cos(xy)=x, find 𝑑𝑦 𝑑π‘₯ βˆ’ csc π‘₯𝑦 βˆ’π‘¦ π‘₯

23 Find the slope of the line tangent to the curve y=arctan(3x) at x = 1 3 .
3 2

24 Day Two Practice

25 Use a right Riemann sum with the four subintervals indicated by the data in the table to approximate π‘Š 𝑑 𝑑𝑑. t (minutes) 5 9 12 20 W(t) degrees F 54.0 58.2 63.1 68.1 70 (70) =

26 Is the previous estimate an overestimate or an underestimate?
It is an overestimate, because the function is always increasing and the right Riemann sum would be above the curve.

27 5 π‘₯ 2 βˆ’2π‘₯𝑦=π‘π‘œπ‘ π‘₯, find 𝑑𝑦 𝑑π‘₯ . 𝑑𝑦 𝑑π‘₯ = βˆ’ sin π‘₯ +2π‘¦βˆ’10π‘₯ βˆ’2π‘₯

28 𝑑𝑦 𝑑π‘₯ = βˆ’π‘₯ 𝑦 Find the solution y = f(x) to the given differential equation with the initial condition f(-1) = 2. 𝑦= 4 π‘₯ 2 +1

29 𝑑𝑦 𝑑π‘₯ = βˆ’π‘₯ 𝑦 2 2 Write an equation for the line tangent to the graph of f at x = -1 if f(-1)=2.
y – 2 = 2(x + 1)

30 Day Three Practice

31 sin⁑(3π‘₯)𝑑π‘₯ = βˆ’cos⁑(3π‘₯) 3 +𝐢

32 lim π‘₯β†’0 2π‘₯ 4 +5 π‘₯ 3 5π‘₯ 4 + 3π‘₯ 3 5 3

33 lim π‘₯β†’0 𝑠𝑖𝑛π‘₯ π‘₯ = 1

34 𝑓 π‘₯ = π‘₯ 2 βˆ’4π‘₯+3π‘“π‘œπ‘Ÿπ‘₯≀2 π‘˜π‘₯+1π‘“π‘œπ‘Ÿπ‘₯>2 The function f is defined above
𝑓 π‘₯ = π‘₯ 2 βˆ’4π‘₯+3π‘“π‘œπ‘Ÿπ‘₯≀2 π‘˜π‘₯+1π‘“π‘œπ‘Ÿπ‘₯>2 The function f is defined above. For what value of k, if any is f continuous at x = 2? k = -1

35 X=0 Where are the minimums? X=-1.5 and x = 6
The function f given by 𝑓 π‘₯ = 1 2 π‘₯ 4 βˆ’3 π‘₯ 3 βˆ’9 π‘₯ 2 has a relative maximum at x = ? X=0 Where are the minimums? X=-1.5 and x = 6

36 What is the slope of the line tangent to the graph of 𝑦= 𝑒 βˆ’π‘₯ π‘₯+3 at x=1?
βˆ’ 5 16𝑒

37 lim β„Žβ†’0 𝑒 3+β„Ž βˆ’ 𝑒 3 β„Ž = 𝑒 3

38 lim π‘₯β†’βˆž 4π‘₯ 6 βˆ’5π‘₯ 3π‘₯ 6 + 7π‘₯ 2 = 4 3

39 𝑑𝑦 𝑑π‘₯ 0 π‘₯ 2 sin 2𝑑 𝑑𝑑= 2π‘₯(𝑠𝑖𝑛 2 π‘₯ 2 )


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