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ページ1:
Review: Limit
lim fey=L.
x+c
informal det
formal def
(p.114~p.115)
old thomas cul
f(x)
lies in here
-x+c (f(c) is not the point.
as x→c,
f>L.
L+E
→
L
•f(x)
2-84
(the
For all X*Xo
in here
↓
discuss)
- point of limit is what's happening on foxs closer and closer to x=c.) (but not f(x) at x=c )
(any)
(L-E, L+E)
We require that for every interval about L, no matter how small,
We can
find
an interval of numbers about to whose f-value all lie
(output)
(x-6x0+8)
D
+
in (L-E, LTE).
X+xo.
Xo
xo-8
the "math"
X0+8
way
of
saying
:
the closer xxo, the closer y=fexs >L.
Definition
We say x²+xo`
lim fax)=L if
given any radius ESO about L
Give It a
<501>
there exists
a radius o about to
such that (for all x that)
。
<|-|<8 implies ₤-L| <=
f(x) does not matter
Try Show that lim (5x-3)=2
We
x
set x0 =1, fix)=5x-3, L=2.
-L<E
can find So such that
we
want
to show that for any ε>0
we
→>>
· | f(x) - L|< E
-
(5-3)-2|<8
oc |x-xo8
04|×-248
Let's find S:
->
| 5x-5|<ε.
which is
5|x-1|<ε.
->
1x-1|< choose S
19/+ (b) > latbl (Triangle inequality).
note: We can also chouse S < & like S = &
can
use the limit def to
We
example show that
x-Xo
then Im [f, (x) + f (x)] = L₁ th₂.
prouf lim theorems.
2ε
fix)-L₁| + | f (x) - (2 | 3 | f(x)=√10)-4-12 | = | fox-L|
If lim fi(x)=4₁ and x mm f2(x)=12
||
@ or 0?
Ⓒ or e
for this to hold: S = min.
proof we wane
to show Vεso, 7870, s.t.
^ { s₁, s₂ }
0</x-x60/98 = | fox)-L| <E!
3°
We now can say x+xo
Jim [f(x)+ fox)] = L
2° from lim f(x)=4 and im fx(x)=LL
Since the criterion holds.
X-9Xo
X-X0
guy.
we know </x-x0|<8 => | f₁00-L₁ | <E
and 0</x-xo/<S₂ => flxs-4/
ページ2:
Review: Continuity 1. fcc) exist The continuity test: Function y = f(x) is continuous (a) the 3 statements are true : 2. im fe exist limit: interior - 2 sided. 见 3. Tim f(x) = f(c). limit value and function value both exist and endpoint-sided equal. prove the following 2 properties, we need IR system knowledge, not in basic cale) Max - min theorem for Continuous Functions If fis continuous on [a, b], then f has Max and min somewhere in [a, b]. That is, for some numbers x1, x2 in [a,b], we have f(x)=m, f(x)=M, and m = f(x) =M at every other pt of [a,b]. The intermediate value theorem for continuous functions A function y = f(x) continuous on [a,b] takes every value between fla) and f(b). f(b) Yo fla). other words, if f(a) = y. < f(b), then yo= f(c) for some ce [a,b]. The IVT implies: ① Connectivity. Root Finding
ページ3:
ReviewDerivative The Derivative of a Function function too) def f(x) = lim f(x+h)-fog) 620 h the difference quotient for f at x: f(x+h)-f(x) h description Differentiable @ a point x called a Differentiable @ a point functim is Differentiable @ every point of Domain a differentiable function then is continuous at x=L. (differentiable), (可微分必連續) Theorem If a function has a derivative at x=L proof To show f is continuous, (at x=C) we need to show 1. f(c) exist. 2. I'm fox) east. XAL > we know for sure! 3. lin fox)=f(c). we know: 1. im (x-c) = I'm f(x)-f(c) 2. XAL X-L 50 xin (x-1) x lim X-91 fox)-f(c) XX-L I'm [(x-1) f(x)-(c)] = I'm [foxy-fis)] = = Rules of Basic Dif. 1. (c)=0 3. (cu)= c du exists = 0xm=0 n-1 > lim [fes- fes] = lim fexs - ft) = 0 4. 1 (u±v) = du = d. * 5. 11 (uv) = du v + udu = u'u+uv' (symmetry) 6. (4) = uv-uv 7. n is negative interger and x#0, de (x²)=nx²+ J RED. ✓74-sub. Chain Rule: = 141 atx at u(x) rewrite: dx Ju. atx. 器 TRIG Dif. (symmetry here) proving. d 1, đc sinx Fest ①def physics: SHM (SMX) 已 (cosk) + 3. dx tanx = Seix 4. Secx = tanxsexx 2. A Cosx = -sinx. 5. ttxe cscx 6. CSC x= - cotxesex sin tacos sin cos tan 平方(相采 Cot tan Sec CSL Sec ☑☑ cot CSC 上乘 對稱加負號 sinx tui xa tan'x + 1 = sec²x cot'x +/ = <sc²x
ページ4:
Other Dif Rules + b (x) = + Q: ? 2 = d =d[en[an] = e en (a) •en(a*) & [en(a*)] e . a. In (a) dx. [x In[a]] a² = a². In (a) The road to MUT: We need to know max and min — relative/absolute (local) e.g. 相對極大vs.絕對板大...... The first derivative theorem for Local Extreme Values (FOT) -max = (a,b)` interior point c of an interval fis defined, If f has local or at and f' is defined at x=C, then filmin proof without works Local max slope o slope so (never positive) (never negative f(c) exisex+c+ lim f(x)-f) = lim f(x)-f(c) != 0 = f(4) = lim fox)-fo X-L x-+- X-L x- Rolle's Theorem (RT). Suppose that y = f(x) is continuous on [a,b] and differentiable in (a,b). If f (a) = f(b)=0 y= f(x) MUT: if f is cont. on [a,b] and dif. in (a,b) there exist in (a,b) S.t. f(b)-f(a) b-a wwwm = slope of secant. =fes. f(s) = slope = M2 slope=m fes a d(x). Let d (x) = f(x)-g(x). We know g(x) = fras + f(b)-f(a) b-a )-f(a) (x-a). → d(x) = foc) - f(a)- (x-a): 'd' = f'-g' f(b)-fla). b-9 b-a d'(x) = f(x)- f(b)-f(a). From RT, = x=c =(a,b) we know for d(x) s.t. d'(c)=0. d' (c) = f(c) - b-a f(63-f(a) =0. f(b)-f(9) f(c)= b-a QED. then there exist at least one c s.t. f(c) = 0. proof €(a,b) we can find extreme values on [9,6] at 1. end points C b 0 = min (f) = f(x) ≤ max(f) = 0 → c is every point in 2. f'co = →one or tho of extreme value 3. f' X exist. * (againse hypothesis) (a,b) not on xa or x=b FDT on x=c, f(c) = extreme f(c)=0
ページ5:
Inverse Trig Functions The Arc Sine y=sin"'x (sin x)=? Let y = sin x < x=sing (sink) = ->> SITE dx = sin x + c. Domain: -1≤x≤| Range: -1/2 =4≤7/2 → do dx = | = cosy √1-X2 - dx = x (sin x) = cosy = ' √1-x2 (-14×41) (-1<*<1) note that (sinx)'= s/nx = cscx + sin x. Sinx ②sin'(-x)=-sin'(x) → odd function. cos'(x) + cos(x) = π | (cos"x)=? ②The Arc Cosine y y=cos" identities: Domain: -1≤x≤1 ② sin" (x) + cos(x) = . Range: Osyst genme me proof | = Lx = = siny Lxx -π x sin'x tan 4=742 y=tan'x Domain:-<x< Range: -7/cy</2 ax tan x =? Y = tan x x = tany dx Y=-72 = Sexy d y=secx • sec'y = 3x+ ST+ x2 dx = tan' x + C -34 " Tu y=secx Domain:1x/>1 Range: Osyst, y4/2 Secy (-1) Secy> √x21 not def. Cotx x=cosy ☑ (45) = - (1x11) = - √√√1-X2 (12x1<1) ·St)ch=105x+C y=cotx y=π a/2 Domain: -00< Range: 0<y<. × (cot x) => ← y=cot"> <x=coty → =|= -css'y ex dy = - (y == sinty = 2x+15 1+x2 S(-1)=='x+C (cat)=- -> 1+x² (sec⭑x)>? -> Sec-x=y x=sexy / tuny secy . tany+1=sec³y - not def. dx tany secy ±√x²/.x :fal = xxx fx-1 12. // TT (IN>1) <> M
ページ6:
(secx) = x/J (sec" (x1) = J ·(-1). (chain rule) √ TX JATT = secx + C = Sec sec |x|+C (x'>1)
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