What is the tangent of f(x)=e^{-3x},(0,1) ?

The tangent of f(x)=e^{-3x},(0,1) is y=-3x+1

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tangent of f(x)=e^{-3x},(0,1)

tangent of

f (x) = e^{- 3 x}, (0, 1)

y = - 3 x + 1

Solution steps

Find the slope of

f (x) = e^{- 3 x} : \frac{df ( x )}{d x} = - 3 e^{- 3 x}

Compute the slope of

f (x) = e^{- 3 x}

(0, 1) : m = - 3

Find the line with slope m=

- 3

and passing through

(0, 1) : y = - 3 x + 1

y = - 3 x + 1

\frac{d x}{d t} = 5 x (6 - x)

\frac{d y}{d x} + 3 y = 50

\frac{d y}{d x} = \frac{y}{x} + \frac{x}{y ^{5}}

\int \frac{12 u}{u ( u - 8 )} d u

k = 0 \sum \infty (k + 1) x^{k}

What is the tangent of f(x)=e^{-3x},(0,1) ?
The tangent of f(x)=e^{-3x},(0,1) is y=-3x+1