What is the equation of the tangent line of f(x)=x−4ex+x2ex−1 at x=4?

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Caroline Carey

2022-03-17

What is the equation of the tangent line of

f (x) = x - \frac{4}{e^{x}} + \frac{x^{2}}{e^{x}} - 1

at x=4?

Pentyrch9ci

Beginner2022-03-18Added 9 answers

Explanation:
Given:

y = f (x) = x - 4 e^{- x} + x^{2} e^{- x} - 1

Evaluate a x=4 so that we know a point on the tangent line

y = f (4) = 4 - 4 e^{- 4} + {(4)}^{2} e^{- 4} - 1

y = 3 + 12 e^{- 4}

The point for the tangent line is

(4, 3 + 12 e^{- 4})

Compute the first derivative:

f^{'} (x) = 4 e^{- x} + 2 x e^{- x} - x^{2} e^{- x}

Evaluate at x=4:

f^{'} (4) = 4 e^{- 4} + 2 (4) e^{- x} - {(4)}^{2} e^{- 4} = - 8 e^{- 4}

The slope

m = - 8 e^{- 4}

Using the point-slope form of the equation of a line:

y - (3 + 12 e^{- 4}) = - 8 e^{- 4} (x - 4)

y - (3 + 12 e^{- 4}) = - 8 e^{- 4} x + 32 e^{- 4}

y = - 8 e^{- 4} x + 3 + 44 e^{- 4}

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