What is the tangent of f(x)=-3e^{-2x}+3 ?

The tangent of f(x)=-3e^{-2x}+3 is y=6e^{-2a_{0}}x-3e^{-2a_{0}}+3-6e^{-2a_{0}}a_{0}

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tangent of f(x)=-3e^{-2x}+3

tangent of

f (x) = - 3 e^{- 2 x} + 3

y = 6 e^{- 2 a_{0}} x - 3 e^{- 2 a_{0}} + 3 - 6 e^{- 2 a_{0}} a_{0}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, - 3 e^{- 2 a_{0}} + 3)

Find the slope of

f (x) = - 3 e^{- 2 x} + 3 : \frac{df ( x )}{d x} = 6 e^{- 2 x}

Compute the slope of

f (x) = - 3 e^{- 2 x} + 3

x = a_{0} : m = 6 e^{- 2 a_{0}}

Find the line with slope m=

6 e^{- 2 a_{0}}

and passing through

(a_{0}, - 3 e^{- 2 a_{0}} + 3) : y = 6 e^{- 2 a_{0}} x - 3 e^{- 2 a_{0}} + 3 - 6 e^{- 2 a_{0}} a_{0}

y = 6 e^{- 2 a_{0}} x - 3 e^{- 2 a_{0}} + 3 - 6 e^{- 2 a_{0}} a_{0}

\int \frac{1}{( x ^{2} + 2 ) ^{\frac{5}{2}}} d x

\int \frac{u - 1}{u ^{2}} d u

\int_{0}^{\infty} x e^{- x^{2}} d x

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

2

What is the tangent of f(x)=-3e^{-2x}+3 ?
The tangent of f(x)=-3e^{-2x}+3 is y=6e^{-2a_{0}}x-3e^{-2a_{0}}+3-6e^{-2a_{0}}a_{0}