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

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

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

tangent of

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

y = 4 x + 3

Solution steps

Find the tangent point:

(- 1, - 1)

Find the slope of

f (x) = 2 x^{3} + 3 x^{2} + 4 x + 2 : \frac{df ( x )}{d x} = 6 x^{2} + 6 x + 4

Compute the slope of

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

x = - 1 : m = 4

Find the line with slope m=

4

and passing through

(- 1, - 1) : y = 4 x + 3

y = 4 x + 3

y^{^{''}} + 3 y^{^{'}} = 9 x

\frac{d}{d x} (e^{(- 5 x^{2} - 2 x^{2})})

\frac{d}{d y} (\frac{y - 2 ( u y )}{y})

y^{^{''}} + 2 y = 18

\frac{1}{s ( s ^{2} + s + 5 )}

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