Let f be the function defined by f(x)= 5cos(pix) - e^(x^2 -4). If g(x)= inverse of f(x), find the equation of the line tangent to the graph if g at point where g(x)=2 I would just like to check that I understood how to do this properly... derivative of f(x) = -5pisin(pix) -2xe^(x^2 -4) Plug in 2 for x -> -4 slopes of inverse functions are reciprocals so slope at g(x)=2 would be -1/4 Then plugging in 2 for x into f(x) because g(x)=2 means f(2)=x So gets point (4,2) in terms of g So I get y= -1/4 (x-4) +2 as the tangent line? :3
@whpalmer4 @sourwing @Luigi0210 @johnweldon1993 help check? :3
@agent0smith @campbell_st @RadEn could you pwease help me check? :3
@zepdrix @robtobey @sourwing
hmmm? I got f'(x) then said slopes of inverse functions are reciprocals
I know that f^-1(x) doesnt equal f'(x) I got f'(x) cuz g'(x) = 1/f'(g(x))
you're basically calculating 1/f'(2)
yes, found f'(x) plugged in x=2 then reciprocal
don't have to do anything by hand, calculator question
You did it right
ok thank you! :D <3
except one minor mistake. it's at (-4,2). not at (4,2)
y= -1/4 (x+4) +2 is the correct answer
why would it be -4 ? http://www.wolframalpha.com/input/?i=5cos%28pix%29+-+e%5E%28x%5E2+-4%29+for+x%3D2
because (2,-4) is the point on f, then this means (-4,2) is the point on g(x)
but wolfram told me that point on f(2) = 4 ;-; http://www.wolframalpha.com/input/?i=5cos%28pix%29+-+e%5E%28x%5E2+-4%29+for+x%3D2
ahh I see what's up now. I took your word for it. You said in the question that you got -4 when x = 2
for derivative :P f'(2) = -4
f(2) = 4 xD
O.O ah crap. I should have read the question carefully :D.
so (4,2) is on g. and 1/f'(f^-1(x)) = 1/f'(2) = 1/-4 = -1/4 yeah it's correct :DD
Thank you ^_^
good job. You understand the concept ^.^b
I hope I do :P APs are in two weeks >.> O_O
ah wait O.O
wut now? o.o stop scaring me ;-; xD
Shouldn't we be calculating 1/f'(4) instead?
oh no nmv XD
;-; meanie why you scare my like dat >,<
nvm XD. This whole derivative of inverse thing can really drive you crazy XDD
lol ikr xD for some reason none of the APCalcBC people know it but apparently the APCalcAB kids are >.>
my grammer just failed me xD
Grammar errors are acceptable in a math class :D
:D yrsh xD
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