What is the tangent of f(x)=4e^{-2x}-9x+1(0.5) ?

The tangent of f(x)=4e^{-2x}-9x+1(0.5) is y=(-8e^{-2a_{0}}-9)x+8a_{0}e^{-2a_{0}}+4e^{-2a_{0}}+0.5

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tangent of f(x)=4e^{-2x}-9x+1(0.5)

tangent of

f (x) = 4 e^{- 2 x} - 9 x + 1 (0.5)

y = (- 8 e^{- 2 a_{0}} - 9) x + 8 a_{0} e^{- 2 a_{0}} + 4 e^{- 2 a_{0}} + 0.5

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 4 e^{- 2 a_{0}} - 9 a_{0} + 0.5)

Find the slope of

f (x) = 4 e^{- 2 x} - 9 x + 1 \cdot 0.5 : \frac{df ( x )}{d x} = - 8 e^{- 2 x} - 9

Compute the slope of

f (x) = 4 e^{- 2 x} - 9 x + 1 \cdot 0.5

x = a_{0} : m = - 8 e^{- 2 a_{0}} - 9

Find the line with slope m=

- 8 e^{- 2 a_{0}} - 9

and passing through

(a_{0}, 4 e^{- 2 a_{0}} - 9 a_{0} + 0.5) : y = (- 8 e^{- 2 a_{0}} - 9) x + 8 a_{0} e^{- 2 a_{0}} + 4 e^{- 2 a_{0}} + 0.5

y = (- 8 e^{- 2 a_{0}} - 9) x + 8 a_{0} e^{- 2 a_{0}} + 4 e^{- 2 a_{0}} + 0.5

4 x - 1

f (x) = \frac{1}{3 x ^{2}}

y - 3 = 5 (x - 2)

f (x) = x + \frac{1}{x}

(- \frac{17}{8}, - \frac{19}{6}), (\frac{25}{6}, \frac{3}{2})

What is the tangent of f(x)=4e^{-2x}-9x+1(0.5) ?
The tangent of f(x)=4e^{-2x}-9x+1(0.5) is y=(-8e^{-2a_{0}}-9)x+8a_{0}e^{-2a_{0}}+4e^{-2a_{0}}+0.5