What is the tangent of f(x)=x^3+2x^2+4x,\at x=-1 ?

The tangent of f(x)=x^3+2x^2+4x,\at x=-1 is y=3x

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tangent of f(x)=x^3+2x^2+4x,\at x=-1

tangent of

f (x) = x^{3} + 2 x^{2} + 4 x, at x = - 1

y = 3 x

Solution steps

Find the tangent point:

(- 1, - 3)

Find the slope of

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

Compute the slope of

f (x) = x^{3} + 2 x^{2} + 4 x

x = - 1 : m = 3

Find the line with slope m=

3

and passing through

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

y = 3 x

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

\int \frac{x ^{- 1} - x ^{- 3} + x ^{- 3}}{x ^{2}} d x

\frac{d}{d x} (\frac{( 3 x ^{2} - 5 )}{( 2 x ^{2} + 1 )})

f (x) = e^{x^{4}}

6 x \frac{d y}{d x} = 2 x e^{x} - 6 y + 6 x^{2}

What is the tangent of f(x)=x^3+2x^2+4x,\at x=-1 ?
The tangent of f(x)=x^3+2x^2+4x,\at x=-1 is y=3x