What is the tangent of f(x)=3e^{-4x},\at x=0 ?

The tangent of f(x)=3e^{-4x},\at x=0 is y=-12x+3

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tangent of f(x)=3e^{-4x},\at x=0

tangent of

f (x) = 3 e^{- 4 x}, at x = 0

y = - 12 x + 3

Solution steps

Find the tangent point:

(0, 3)

Find the slope of

f (x) = 3 e^{- 4 x} : \frac{df ( x )}{d x} = - 12 e^{- 4 x}

Compute the slope of

f (x) = 3 e^{- 4 x}

x = 0 : m = - 12

Find the line with slope m=

- 12

and passing through

(0, 3) : y = - 12 x + 3

y = - 12 x + 3

\frac{s}{s ^{2} - 2 s + 10}

\frac{d}{d x} (\frac{6 x ^{2} + 4 x + 2}{x})

\frac{d}{d y} (\frac{y ^{4} + 48}{24 y})

\int \frac{x ^{3} - 1}{x ^{4} - 4 x} d x

\frac{\partial}{\partial x} (\frac{1}{\frac{1}{x} + \frac{1}{y}})

What is the tangent of f(x)=3e^{-4x},\at x=0 ?
The tangent of f(x)=3e^{-4x},\at x=0 is y=-12x+3