What is the tangent of f(x)=3-e^{-3x},\at x=-ln(2) ?

The tangent of f(x)=3-e^{-3x},\at x=-ln(2) is y=24x-5+24ln(2)

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tangent of f(x)=3-e^{-3x},\at x=-ln(2)

tangent of

f (x) = 3 - e^{- 3 x}, at x = - ln (2)

y = 24 x - 5 + 24 ln (2)

Solution steps

Find the tangent point:

(- ln (2), - 5)

Find the slope of

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

Compute the slope of

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

x = - ln (2) : m = 24

Find the line with slope m=

24

and passing through

(- ln (2), - 5) : y = 24 x - 5 + 24 ln (2)

y = 24 x - 5 + 24 ln (2)

\frac{\partial}{\partial x} (20 + 5 sin (x) cos (x - y))

\int \frac{e ^{2 x}}{x} d x

\frac{d}{d x} (\frac{3}{x ^{2} - 1})

\int sin (3 x) sin (7 x) d x

What is the tangent of f(x)=3-e^{-3x},\at x=-ln(2) ?
The tangent of f(x)=3-e^{-3x},\at x=-ln(2) is y=24x-5+24ln(2)