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

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

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

tangent of

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

y = 2 x + 3

Solution steps

Find the tangent point:

(1, 5)

Find the slope of

y = 3 + 4 x^{2} - 2 x^{3} : \frac{d y}{d x} = - 6 x^{2} + 8 x

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = 2

Find the line with slope m=

2

and passing through

(1, 5) : y = 2 x + 3

y = 2 x + 3

n = 0 \sum \infty \frac{2 n}{n !}

\int (e^{9 x} + 7 e^{- x}) d x

\frac{\partial}{\partial x} (- 2 x^{4} - 3 y^{3})

\int \frac{4 x - 2}{x ( x + 1 ) ( x - 2 )} d x

\int_{4}^{\infty} e^{- 7 x} d x

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