anonymous
  • anonymous
find the linear approximation L(x) of the function f(x)=cos(9x) at a=pi/2 is L(x)=A+Bx
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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bahrom7893
  • bahrom7893
okay workin on it
bahrom7893
  • bahrom7893
L(x) = f(a) + f'(a)(x - a)
bahrom7893
  • bahrom7893
f'(x) = -Sin(9x) * 9

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bahrom7893
  • bahrom7893
f'(pi/2) = -9Sin(9pi/2) = -9*1 = -9
bahrom7893
  • bahrom7893
And f(pi/2) = Cos(9pi/2) = 0
anonymous
  • anonymous
ok whats A and what B
bahrom7893
  • bahrom7893
L(x) = 0 - 9(x-pi/2) L(x) = A + Bx (<--was this right, i keep forgetting the format?)
anonymous
  • anonymous
yes but i do not know how u got ur answer at all
bahrom7893
  • bahrom7893
Oh okay so let me explain... L(x) = f(a) + f'(a)(x - a) <-Linear approximation formula
anonymous
  • anonymous
i got that
bahrom7893
  • bahrom7893
Now for that you are going to need: f(a), f'(a) and x-a
bahrom7893
  • bahrom7893
f(a) = f(pi/2)
bahrom7893
  • bahrom7893
f(pi/2)=cos(9[pi/2]) = 0
anonymous
  • anonymous
i got that the cos (9pi/2)= a decimal
bahrom7893
  • bahrom7893
what is ur calculator's mode in? I bet it's in degrees.. supposed to be in radians
bahrom7893
  • bahrom7893
pi/2 is a radian measure, not degree measure..
bahrom7893
  • bahrom7893
try again? change the mode and find Cos(9pi/2)
anonymous
  • anonymous
i am in radians
bahrom7893
  • bahrom7893
okay can u go to mode and type in the values u have in here?
anonymous
  • anonymous
dont kno how
bahrom7893
  • bahrom7893
what calculator are u using?
anonymous
  • anonymous
ti 83
bahrom7893
  • bahrom7893
okay press MODE
bahrom7893
  • bahrom7893
now type whatever u see in that screen here
bahrom7893
  • bahrom7893
MODE is right next to 2nd key
anonymous
  • anonymous
i see stuff like normal Sci Eng FLoat 0123456789 Radia Degree Func PAr Pol Seq
bahrom7893
  • bahrom7893
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
anonymous
  • anonymous
dont know how to do that either
bahrom7893
  • bahrom7893
do u have a camera?
bahrom7893
  • bahrom7893
wait nevermind, trust me Cos(9pi/2) is 0
bahrom7893
  • bahrom7893
okay?
anonymous
  • anonymous
ok
bahrom7893
  • bahrom7893
L(x) = f(a) + f'(a)(x - a) <-Linear approximation formula f(a) = f(pi/2) f(pi/2)=cos(9[pi/2]) = 0
anonymous
  • anonymous
ok
bahrom7893
  • bahrom7893
f'(x) = [cos(9x)]' = USE CHAIN RULE = -Sin(9x) * 9 = -9Sin(9x)
bahrom7893
  • bahrom7893
f'(pi/2) = -9 Sin(9pi/2) = -9 * 1 = -9
bahrom7893
  • bahrom7893
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)
bahrom7893
  • bahrom7893
L(pi/2) = -9x - 9pi/2
bahrom7893
  • bahrom7893
L(pi/2) = -9x - 9pi/2 L(x) = A + Bx L(pi/2) = -9pi/2 - 9x
bahrom7893
  • bahrom7893
A = -9pi/2 B = -9
bahrom7893
  • bahrom7893
Check the answer
bahrom7893
  • bahrom7893
carra u here? I made a mistake, -9 (x - pi/2) is not -9pi/2 - 9x, its 9pi/2 - 9x
bahrom7893
  • bahrom7893
Double negative is a positive... -9 * -pi/2 is 9pi/2 not -9pi/2

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