anonymous
  • anonymous
Calculus: Find an equation of the tangent line to the curve at the point (1,1/e). y=x^4*e^-x My attempt: y'=(4x^3*e^-x)-(x^4*e^-x) y-y1=m(x-x1) y-(1/e)=(4x^3)(e^-x)-((x^4)(e^-x))(x-1) y=4x^4 e^-x - x^5 e^-x - 4x^3 e^-x + x^4 e^-x +1/e What did I do wrong? I'm reviewing for a test and need a refresher, thanks.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
y'=(4x^3*e^-x)-(x^4*e^-x) x=1 y'=4/e-1/e=3/e this is your slope..
anonymous
  • anonymous
i think i see the problem with your solution remember that the slope is a number, not a fromula to find the slope take the derivative and then replace x by the specific number you have
anonymous
  • anonymous
3/e=(y-1/e)/(x-1) y=(3x/e)-2

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anonymous
  • anonymous
oh okay that makes sense. thanks!
anonymous
  • anonymous
\[f'(x)=4x^3e^{-x}-x^4e^{-x}\] \[m=f'(1)=4e^{-1}-e^{-1}=\frac{3}{e}\]as cinar wrote. that number is your slope, not the derivative
anonymous
  • anonymous
this is your tangent eq.. \[y=\frac{3}{e}x-2\]
anonymous
  • anonymous
* y=3x/e - 2/e

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