What is the tangent of f(x)=3x^2-5x-1+e^{-x},\at x=0 ?

The tangent of f(x)=3x^2-5x-1+e^{-x},\at x=0 is y=-6x

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tangent of f(x)=3x^2-5x-1+e^{-x},\at x=0

tangent of

f (x) = 3 x^{2} - 5 x - 1 + e^{- x}, at x = 0

y = - 6 x

Solution steps

Find the tangent point:

(0, 0)

Find the slope of

f (x) = 3 x^{2} - 5 x - 1 + e^{- x} : \frac{df ( x )}{d x} = 6 x - e^{- x} - 5

Compute the slope of

f (x) = 3 x^{2} - 5 x - 1 + e^{- x}

x = 0 : m = - 6

Find the line with slope m=

- 6

and passing through

(0, 0) : y = - 6 x

y = - 6 x

\int 2 x^{2} - 10 x d x

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\frac{d ^{2}}{d x ^{2}} (\frac{8}{9 x ^{5}})

\frac{\partial}{\partial z} (\frac{x - y}{y + z})

\int (e^{t} + t^{e}) d t

What is the tangent of f(x)=3x^2-5x-1+e^{-x},\at x=0 ?
The tangent of f(x)=3x^2-5x-1+e^{-x},\at x=0 is y=-6x