Notes: Linear Approximation Day 6 Application of Derivatives Classwork Example One: Find the linearization L(x) of the function at a.
A. 22, 5,
B. , 4
, Example Two: Compute the linearization of √−1 1 Example Three: Approximate using linearization. A. √17 B. ln 1.07 C. 1.999 3 D.) sin 89°
Tangent line at *s Y - Yi -
- MH- X ,) Point f-(57=-245.5-2125) y -1-45=-191×-5 )
Slope f- ' (57--191-415) y= - 191×-5) - 45
f- '1×7=1 - 4X
argent line at * I, y - I =2lX - If )
PointH'E)=L tan y=2( x - 'E)tl Slope f- 'CE,)=2_csectif.EE/=Iflxl--tanxfYxl..=sec2x Tangent line at # I flared
- '=CDeEClXD=I y - 1=21×-1) Point f-(17=-1 FIX) -_ x''2.ex- I 2
Slope f- ' (D= -312 fy×keF! + e
' !zx- 'k y -
zrx
_*⇒
Tangent line at x=l6 y= (x- 161+4 y24. 125 the tangent line lies •←←#→ Point 'fCl67=-4 - ABIE f-(x) so OUR ,
Slope f'Clo)-18--26=2743 y - - tool" -167+4 Th -- 4.123 approximation is an
' 6
f- '
f- (x)=X3 y -18=121×-12) Approximation A
iota:&.hn#ci.nze.Estx---2yy:.YfEt?ahaIg.s ( same
Slope f 't-27=-1231-25 y= . 012-8 C-1.99913=-7.988005 µ f- ' CX) -- 3×2 yay -7,988
L
FIX) -- Sino y - 1=04-IL) convert 897rad ( ←•→Tangent line at # Iz y=o(x-E)ti 890 .
L yer Point HE)=l_ Sin'E 890=81# same
Slope f- ' C'E) =-0 COSTE Y -- 014¥ - 9¥)H ego
Approximation f 'Cx7= cost y -- Offgto) t ' yurt
921 silnfgqo ) I.99984
Notes: Linear Approximation Day 6 Application of Derivatives Example 4: How would you see it on the AP exam? A. The function is twice differentiable with 2 1, 2 4, and 2 3. What is the value of the approximation of 1.9 using the line tangent to the graph of at 2?
B. Let be a differentiable function such that 3 2 and 3 5. If the tangent line to the graph of at 3 is used to find an approximation to a zero of , that approximation is
1-4: Find the linearization L(x) of the function at a. 1. 4 2( ) 3 , 1f x x x a= + = 2. ( ) sin ,
6 f x x a= =
3. ( ) , 4f x x a= = 4. 3 4( ) , 16f x x a= =
(A) 0.4 (B) 0.6 (C) 0.7 (D) 1.3 (E) 1.4
(A) 0.4 (B) 0.5 (C) 2.6 (D) 3.4 (E) 5.5
@8apno9ehftfignzeIHHatx-2rfHlzt-3-pos9sie.t:¥¥.ie#ii:iii:ihEy---.4ttyz.6isanY2.6 under approximation
because f- ' ' CX ) >0 .
Tangent line of fix) atx-- 3 y 51×-37+2 f- (37=2 zerolrootlxiinterceptl solution@ f' ( 31=5 HI , @ D
y - 2=51×-37 0=5×-15+2 13=5X x -_ Eg -_ 245--2.6
5. Find the linear approximation of the function ( ) 1f x x= at a=0 and use it to approximate
the numbers 0.9 and 0.99 . 6-10: Use a linear approximation (or differentials) to estimate the given number.
6. ( )41.999
7. 0.015e
2 6 2 y x= +
3) ( )1 4 2 4
y x= +
.99 .995
6) ( )41.999 15.968 7) 0.015 .985e 8) 3 1001 10.003 9) 1 .249875
4.002 10) ( )tan 44 .965 11) 99.8 9.99