Question

find the linear approximation L(x) of the function f(x)=cos(9x) at a=pi/2 is L(x)=A+Bx

Answer 1

okay workin on it

Answer 2

L(x) = f(a) + f'(a)(x - a)

Answer 3

f'(x) = -Sin(9x) * 9

Answer 4

f'(pi/2) = -9Sin(9pi/2) = -9*1 = -9

Answer 5

And f(pi/2) = Cos(9pi/2) = 0

Answer 6

ok whats A and what B

Answer 7

L(x) = 0 - 9(x-pi/2)
L(x) = A + Bx (<--was this right, i keep forgetting the format?)

Answer 8

yes but i do not know how u got ur answer at all

Answer 9

Oh okay so let me explain...
L(x) = f(a) + f'(a)(x - a) <-Linear approximation formula

Answer 10

i got that

Answer 11

Now for that you are going to need:
f(a), f'(a) and x-a

Answer 12

f(a) = f(pi/2)

Answer 13

f(pi/2)=cos(9[pi/2]) = 0

Answer 14

i got that the cos (9pi/2)= a decimal

Answer 15

what is ur calculator's mode in? I bet it's in degrees.. supposed to be in radians

Answer 16

pi/2 is a radian measure, not degree measure..

Answer 17

try again? change the mode and find Cos(9pi/2)

Answer 18

i am in radians

Answer 19

okay can u go to mode and type in the values u have in here?

Answer 20

dont kno how

Answer 21

what calculator are u using?

Answer 22

ti 83

Answer 23

okay press MODE

Answer 24

now type whatever u see in that screen here

Answer 25

MODE is right next to 2nd key

Answer 26

i see stuff like normal Sci Eng
FLoat 0123456789
Radia Degree
Func PAr Pol Seq

Answer 27

what are the values, wait nevermind, can u take a picture of that mode screen and upload it on tinypic.com ? no registration needed btw

Answer 28

dont know how to do that either

Answer 29

do u have a camera?

Answer 30

wait nevermind, trust me Cos(9pi/2) is 0

Answer 31

okay?

Answer 32

ok

Answer 33

L(x) = f(a) + f'(a)(x - a) <-Linear approximation formula
f(a) = f(pi/2)
f(pi/2)=cos(9[pi/2]) = 0

Answer 34

ok

Answer 35

f'(x) = [cos(9x)]' = USE CHAIN RULE = -Sin(9x) * 9 = -9Sin(9x)

Answer 36

f'(pi/2) = -9 Sin(9pi/2) = -9 * 1 = -9

Answer 37

L(x) = f(a) + f'(a)(x - a) <-Linear approximation formula
f(a) = f(pi/2)
f(pi/2)=cos(9[pi/2]) = 0
f'(pi/2) = -9 Sin(9pi/2) = -9 * 1 = -9
L(x) = f(a) + f'(a)(x - a)
L(pi/2) = 0 + -9 (x - pi/2)

Answer 38

L(pi/2) = -9x - 9pi/2

Answer 39

L(pi/2) = -9x - 9pi/2
L(x) = A + Bx
L(pi/2) = -9pi/2 - 9x

Answer 40

A = -9pi/2
B = -9

Answer 41

Check the answer

Answer 42

carra u here?
I made a mistake, -9 (x - pi/2) is not -9pi/2 - 9x, its 9pi/2 - 9x

Answer 43

Double negative is a positive... -9 * -pi/2 is 9pi/2 not -9pi/2