What is the tangent of f(x)=e^x(8+3x+5x^2),\at x=0 ?

The tangent of f(x)=e^x(8+3x+5x^2),\at x=0 is y=11x+8

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tangent of f(x)=e^x(8+3x+5x^2),\at x=0

tangent of

f (x) = e^{x} (8 + 3 x + 5 x^{2}), at x = 0

y = 11 x + 8

Solution steps

Find the tangent point:

(0, 8)

Find the slope of

f (x) = e^{x} (8 + 3 x + 5 x^{2}) : \frac{df ( x )}{d x} = 5 e^{x} x^{2} + 13 e^{x} x + 11 e^{x}

Compute the slope of

f (x) = e^{x} (8 + 3 x + 5 x^{2})

x = 0 : m = 11

Find the line with slope m=

11

and passing through

(0, 8) : y = 11 x + 8

y = 11 x + 8

\frac{d}{d x} (x^{3} e^{- 4 x})

\frac{d y}{d x} + y - 4 cos (x) = 0

\int_{0}^{5} \frac{1}{x ^{\frac{17}{18}}} d x

\int v e^{uv} d u

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

What is the tangent of f(x)=e^x(8+3x+5x^2),\at x=0 ?
The tangent of f(x)=e^x(8+3x+5x^2),\at x=0 is y=11x+8