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

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

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tangent of y=3-2x+4x^3,\at x=-1

tangent of

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

y = 10 x + 11

Solution steps

Find the tangent point:

(- 1, 1)

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}

x = - 1 : m = 10

Find the line with slope m=

10

and passing through

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

y = 10 x + 11

\frac{d}{d x} (ln (x^{16}))

y^{^{'}} = x y + x + y + 1

\int_{0}^{1} 7 (x^{2} + 1) e^{- x} d x

\int \frac{x ^{3} - 8 x}{x ^{2}} d x

\frac{d}{d x} ((x^{2} + sin (x)) sec (x))

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