MTALHAHASSAN2
  • MTALHAHASSAN2
find the point on the curve f (x)=(x^2-1)^2 for which the tangent line is horizontal
Mathematics
jamiebookeater
  • jamiebookeater
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IrishBoy123
  • IrishBoy123
\(f = (x^2-1)^2\) you learning calculus?
MTALHAHASSAN2
  • MTALHAHASSAN2
yes
IrishBoy123
  • IrishBoy123
so you know to differentiate f(x) and set \(f'(x) = 0\) ??

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MTALHAHASSAN2
  • MTALHAHASSAN2
yeah i do
MTALHAHASSAN2
  • MTALHAHASSAN2
wait let be show you then
IrishBoy123
  • IrishBoy123
cool!
MTALHAHASSAN2
  • MTALHAHASSAN2
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MTALHAHASSAN2
  • MTALHAHASSAN2
right?
IrishBoy123
  • IrishBoy123
nah \(f = (x^2-1)^2\) \(f' = 2(x^2-1)(2x)=\dots\) if that is confusing, let \(u = x^2 -1 \) and use \(\frac{dy}{dx} = \frac{dy}{du} \frac{du}{dx}\)
MTALHAHASSAN2
  • MTALHAHASSAN2
then is it be 4x(x^2-1)
IrishBoy123
  • IrishBoy123
yes let's do it long way too \(f = (x^2-1)^2 = x^4 - 2 x^2 + 1\) \(f' = 4x^3 - 4x = 4x(x^2-1)\) \(\checkmark\)
IrishBoy123
  • IrishBoy123
|dw:1443861295810:dw|
MTALHAHASSAN2
  • MTALHAHASSAN2
oh ok got you
MTALHAHASSAN2
  • MTALHAHASSAN2
thnx a lot so the 4x(x^2-1) be the tangent line is
IrishBoy123
  • IrishBoy123
\(f' = 0 \implies 4x(x^2-1) = 0\) solve for x, you will get 3 turning points
IrishBoy123
  • IrishBoy123
|dw:1443862179850:dw|

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