What is the tangent of f(x)=(x^3-4x+1)e^x,\at x=0 ?

The tangent of f(x)=(x^3-4x+1)e^x,\at x=0 is y=-3x+1

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tangent of f(x)=(x^3-4x+1)e^x,\at x=0

tangent of

f (x) = (x^{3} - 4 x + 1) e^{x}, at x = 0

y = - 3 x + 1

Solution steps

Find the tangent point:

(0, 1)

Find the slope of

f (x) = (x^{3} - 4 x + 1) e^{x} : \frac{df ( x )}{d x} = e^{x} x^{3} + 3 e^{x} x^{2} - 4 e^{x} x - 3 e^{x}

Compute the slope of

f (x) = (x^{3} - 4 x + 1) e^{x}

x = 0 : m = - 3

Find the line with slope m=

- 3

and passing through

(0, 1) : y = - 3 x + 1

y = - 3 x + 1

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What is the tangent of f(x)=(x^3-4x+1)e^x,\at x=0 ?
The tangent of f(x)=(x^3-4x+1)e^x,\at x=0 is y=-3x+1