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

The tangent of f(x)=2+2e^{-x},\at x=ln(5) is y=-2/5 x+12/5+2/5 ln(5)

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tangent of f(x)=2+2e^{-x},\at x=ln(5)

tangent of

f (x) = 2 + 2 e^{- x}, at x = ln (5)

y = - \frac{2}{5} x + \frac{12}{5} + \frac{2}{5} ln (5)

Solution steps

Find the tangent point:

(ln (5), \frac{12}{5})

Find the slope of

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

Compute the slope of

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

x = ln (5) : m = - \frac{2}{5}

Find the line with slope m=

- \frac{2}{5}

and passing through

(ln (5), \frac{12}{5}) : y = - \frac{2}{5} x + \frac{12}{5} + \frac{2}{5} ln (5)

y = - \frac{2}{5} x + \frac{12}{5} + \frac{2}{5} ln (5)

\frac{1}{6 x ^{3}}

\int sin (x) cos^{4} (x) d x

81 y^{^{''}} + 180 y^{^{'}} + 100 y = 0

\int_{0}^{20} 3 x d x

\int \frac{1}{t ^{2} 100 - t ^{2}} d t

What is the tangent of f(x)=2+2e^{-x},\at x=ln(5) ?
The tangent of f(x)=2+2e^{-x},\at x=ln(5) is y=-2/5 x+12/5+2/5 ln(5)