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Calculus notebook with exercises
Typology: Exercises
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I (a) y = 4x + 9x outside (^) -> In inside ->^ 4x+ 922 (b)y
an(n2+ 1) outside ->^ tanz inside ->n+ 1 (2) y = secm outside (^) - i I inside ->^ sec u (d) y^
(
ex)
outside ->^ i inside
186 (see f(x) =^ ((n
A(x +^ 1) #y =^3 (Ay
tH. An+1+y = 3Ay
13x-y-1. A(x
· Find an (^) equation for the^ tangent lineat (^1) , ) y = f'(a)(n
3ny y- I
x x- 3 y = Sy
n) n2 + yz = y
y = x - x y(y
= x(1 - x)
. f(x) g()
· Me M ↓o (^) Jr. W
i ginegist
I ↓ I IAL
-^ I & I I 'I' : I 1 I 1 I
Absolut maximum (maximo^ absoluto) ·Extreme values (Absolut (^) minimum (^) (minima relatio
(^3) - (^2) - (^1) -
problem withthisspeedometer,"fance^ complained to^ the^ clerk. "Yesterday Icycled the^ 22-mileRogadzo Road^ Trailin^78 minutes. and notonce^ didthe^ speedometer read^ above^15 milesper hour!" "Yeah?"responded the^ clock.^ "That's^ the^ problem?"How^ didLance use the^ Mean^ Value^ Theorem to^ explainhiscomplaint? s =^22 miles^ #15miles/h t = 78 minutes v =^ S^ -^ -> E v = Mut: If I^ iscontinuous on^ Ia, b) and (^) differentiable on (^) (a, b), then^ there^ existsatleastone value a in (^) (a, (^) b) such that f(x) = 1 = f(b) - f(a) = f()(b
(^15) miles/per how
growth of a^ sunflower^ during the first (^100) days after sprouting^ is modeled^ well^ by the log istic^ case^ y:h(t).^ Estimate^ the growth
the (^) point (^) of inflection and (^) explainits significance. Then make a rough sketch (^) of the^ just and sewerd h dairatives of
D (^) h(t) (^250) - Near the^ point of inflection,^ the (^) curre (^200) - 140,150) is roughly a straight
(^130) -------.(((r) =^0 through (33,200) and (^) (33,100) (^100) - "
e (^30) - I (^) and so (^) ob Ibot 20
& D Rate of change:
⑳
t SECTION 4.3^ ~ In (^) Exercises15-36, (^) find the (^) transition (^) points, intervals (^) of increase/ decrease, (^) concarity, and^ asymptotic behavior.^ Then^ sketch^ the^ graph, withthis informationindicated. 27.y =^ x^ -^ 4F^ ·^ Ata = 0 the (^) derivativedoes not exist y
=
F D(9)^ = [1, + 3) y
2 =^0 = ) = (^1) =) (^) F(2)-=) m =4. 1 4
I
D(y) = (0, + 0) to^ him^ f(n) =^ - > Asintotavertical x-^ -^ D
2a(y" = 0 = 31 - x =^0 = 11 - 5 =
7 (0,
conce up = 1 - 2 = 0 = 31 - F = E = E = 0 = r= 0 = (n
......
n - -^ - tim (^) f(x) =^0 Asintota horizontal n -^ i
I ((n) =^ ( -( +x2) 4 19:f'(a)^
(v +a) eri-Tasz en- ?(t) e -1 -^ 12 (a) (1+2) (^) nFr
By = - 5.0.3=-0,03125.cal-10/
=- 3,13 +^ 3,125^ =^ -0,
y'(a) De A t^ h = 20km, P (^) =5.3 (^) kilopascals. ·- f) I Estimate^ P^ at^ altitude^ 20.3^ km.
a
20.3) th
24(25) 3D R = 10. 3 = 4 πr2 (^) S = 4 π(25) If R^ =^25 then, xf^ = y'(a)Bm
E As =^ S'(23)Ar danuts