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

The tangent of y=3-2x+4x^3,(-1,1) is y=10x+11

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tangent of y=3-2x+4x^3,(-1,1)

tangent of

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

y = 10 x + 11

Solution steps

Find the slope of

y = 3 - 2 x + 4 x^{3} : \frac{d y}{d x} = 12 x^{2} - 2

Compute the slope of

y = 3 - 2 x + 4 x^{3}

(- 1, 1) : m = 10

Find the line with slope m=

10

and passing through

(- 1, 1) : y = 10 x + 11

y = 10 x + 11

\int 1 + cos (x) d x

x^{2} \frac{d y}{d x} - 2 x^{3} e^{- \frac{1}{x}} = y

\int sin (5 x) sin (4 x) d x

(4.1) (8.2)

\frac{d}{d x} (\frac{sec ( x ) + csc ( x )}{csc ( x )})

What is the tangent of y=3-2x+4x^3,(-1,1) ?
The tangent of y=3-2x+4x^3,(-1,1) is y=10x+11