Question

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.

Answer 1

y'=(4x^3*e^-x)-(x^4*e^-x)
x=1
y'=4/e-1/e=3/e this is your slope..

Answer 2

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

Answer 3

3/e=(y-1/e)/(x-1)

y=(3x/e)-2

Answer 4

oh okay that makes sense. thanks!

Answer 5

\[f'(x)=4x^3e^{-x}-x^4e^{-x}\]
\[m=f'(1)=4e^{-1}-e^{-1}=\frac{3}{e}\]as cinar wrote. that number is your slope, not the derivative

Answer 6

this is your tangent eq..
\[y=\frac{3}{e}x-2\]

Answer 7

* y=3x/e - 2/e