What is the tangent of f(x)=e^{-5x},\at x=0 ?

The tangent of f(x)=e^{-5x},\at x=0 is y=-5x+1

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tangent of f(x)=e^{-5x},\at x=0

tangent of

f (x) = e^{- 5 x}, at x = 0

y = - 5 x + 1

Solution steps

Find the tangent point:

(0, 1)

Find the slope of

f (x) = e^{- 5 x} : \frac{df ( x )}{d x} = - 5 e^{- 5 x}

Compute the slope of

f (x) = e^{- 5 x}

x = 0 : m = - 5

Find the line with slope m=

- 5

and passing through

(0, 1) : y = - 5 x + 1

y = - 5 x + 1

\frac{d}{d s} (\frac{s ^{2} - 8 s + 15}{s ^{2} + 3 s + 2})

2 y y^{^{'}} + 4 = y^{2} + 4 x

x (7 x + 3)^{\frac{1}{3}}

\int_{0}^{1} x^{3} (1 + x^{4})^{3} d x

ln (x + 2)

What is the tangent of f(x)=e^{-5x},\at x=0 ?
The tangent of f(x)=e^{-5x},\at x=0 is y=-5x+1