Mathematics: 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^-*e^-x) y-y1=m(x-x1) y-(1/e)=(4x^3)(e^-)(e^--1) y=4x^4 e^-x - x^5 e^-x - 4x^3 e^-x + x^4 e^-x +1/e

Find an equation of the tangent line to the curve

Ответ:

ella5916

18.01.2022

We will use y-y1=m(x-x1), where m= f '(x)= 4x^3e^-x - x^4e^-x, f ' (1) =3e^-1
f (1) =4e^-1
so the equation of tangent line tothe curve at the point (1,1/e) is
y= 3e^-1(x-1) + 4e^-1, or y= 3e^-1(x)) + e^-1,

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Ответ:

sparkybig12

08.12.2022

Mathematics

4. Vertical angles theorem

6. 3x + 4 =49

Step-by-step explanation:

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Find an equation of the tangent line to the curve at the point (1,1/e). y=x^4e^-x
my attempt:
y'=(4x^3e^-*e^-x)
y-y1=m(x-x1)
y-(1/e)=(4x^3)(e^-)(e^--1)
y=4x^4 e^-x - x^5 e^-x - 4x^3 e^-x + x^4 e^-x +1/e

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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^-*e^-x) y-y1=m(x-x1) y-(1/e)=(4x^3)(e^-)(e^--1) y=4x^4 e^-x - x^5 e^-x - 4x^3 e^-x + x^4 e^-x +1/e

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Find an equation of the tangent line to the curve at the point (1,1/e). y=x^4e^-x
my attempt:
y'=(4x^3e^-*e^-x)
y-y1=m(x-x1)
y-(1/e)=(4x^3)(e^-)(e^--1)
y=4x^4 e^-x - x^5 e^-x - 4x^3 e^-x + x^4 e^-x +1/e