What is the tangent of f(x)=5e^{4x}+e^{-x^2},\at x=1 ?

The tangent of f(x)=5e^{4x}+e^{-x^2},\at x=1 is y=(20e^4-2/e)x-15e^4+3/e

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tangent of f(x)=5e^{4x}+e^{-x^2},\at x=1

tangent of

f (x) = 5 e^{4 x} + e^{- x^{2}}, at x = 1

y = (20 e^{4} - \frac{2}{e}) x - 15 e^{4} + \frac{3}{e}

Solution steps

Find the tangent point:

(1, 5 e^{4} + \frac{1}{e})

Find the slope of

f (x) = 5 e^{4 x} + e^{- x^{2}} : \frac{df ( x )}{d x} = 20 e^{4 x} - 2 e^{- x^{2}} x

Compute the slope of

f (x) = 5 e^{4 x} + e^{- x^{2}}

x = 1 : m = 20 e^{4} - \frac{2}{e}

Find the line with slope m=

20 e^{4} - \frac{2}{e}

and passing through

(1, 5 e^{4} + \frac{1}{e}) : y = (20 e^{4} - \frac{2}{e}) x - 15 e^{4} + \frac{3}{e}

y = (20 e^{4} - \frac{2}{e}) x - 15 e^{4} + \frac{3}{e}

f (x) = 6 x^{2}

24 y^{^{''}} + 13 y^{^{'}} - 2 y = 0

(6, - 8), (- 2, 7)

\int - csc (5 x) cot (5 x) d x

x \to 4 - lim (\frac{2 - x}{x - 4})

What is the tangent of f(x)=5e^{4x}+e^{-x^2},\at x=1 ?
The tangent of f(x)=5e^{4x}+e^{-x^2},\at x=1 is y=(20e^4-2/e)x-15e^4+3/e