What is the tangent of f(x)=7e^{-0.03x},\at x=9 ?

The tangent of f(x)=7e^{-0.03x},\at x=9 is y=-0.16030…x+7/(e^{0.27)}+1.44278…

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tangent of f(x)=7e^{-0.03x},\at x=9

tangent of

f (x) = 7 e^{- 0.03 x}, at x = 9

y = - 0.16030 \dots x + \frac{7}{e ^{0.27}} + 1.44278 \dots

Solution steps

Find the tangent point:

(9, \frac{7}{e ^{0.27}})

Find the slope of

f (x) = 7 e^{- 0.03 x} : \frac{df ( x )}{d x} = - 0.21 e^{- 0.03 x}

Compute the slope of

f (x) = 7 e^{- 0.03 x}

x = 9 : m = - 0.16030 \dots

Find the line with slope m=

- 0.16030 \dots

and passing through

(9, \frac{7}{e ^{0.27}}) : y = - 0.16030 \dots x + \frac{7}{e ^{0.27}} + 1.44278 \dots

y = - 0.16030 \dots x + \frac{7}{e ^{0.27}} + 1.44278 \dots

y = \frac{( 3 x - 9 )}{( 2 x + 8 )}

\int (x^{3} + x) e^{- 4 x} d x

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

\frac{d}{d x} (2 (3 - x^{2})^{5})

\int_{0}^{3} 2 π x (9 - x^{2}) d x

What is the tangent of f(x)=7e^{-0.03x},\at x=9 ?
The tangent of f(x)=7e^{-0.03x},\at x=9 is y=-0.16030…x+7/(e^{0.27)}+1.44278…