What is the tangent of f(x)=8e^{-0.5x},\at x=7 ?

The tangent of f(x)=8e^{-0.5x},\at x=7 is y=-0.12078…x+8/(e^{3.5)}+0.84552…

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tangent of f(x)=8e^{-0.5x},\at x=7

tangent of

f (x) = 8 e^{- 0.5 x}, at x = 7

y = - 0.12078 \dots x + \frac{8}{e ^{3.5}} + 0.84552 \dots

Solution steps

Find the tangent point:

(7, \frac{8}{e ^{3.5}})

Find the slope of

f (x) = 8 e^{- 0.5 x} : \frac{df ( x )}{d x} = - 4 e^{- 0.5 x}

Compute the slope of

f (x) = 8 e^{- 0.5 x}

x = 7 : m = - 0.12078 \dots

Find the line with slope m=

- 0.12078 \dots

and passing through

(7, \frac{8}{e ^{3.5}}) : y = - 0.12078 \dots x + \frac{8}{e ^{3.5}} + 0.84552 \dots

y = - 0.12078 \dots x + \frac{8}{e ^{3.5}} + 0.84552 \dots

\frac{\partial}{\partial x} (4 x sin (x y))

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

x \to 2 lim ((\frac{x + 3}{2 x + 1})^{\frac{2}{x - 2}})

\frac{d}{d x} (x arccos (3 x))

\int_{- 2}^{4} \frac{1}{( x - 3 ) ^{3}} d x

What is the tangent of f(x)=8e^{-0.5x},\at x=7 ?
The tangent of f(x)=8e^{-0.5x},\at x=7 is y=-0.12078…x+8/(e^{3.5)}+0.84552…