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

The tangent of x^4-4x^3-2,\at x=1 is y=-8x+3

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

tangent of

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

y = - 8 x + 3

Solution steps

Find the tangent point:

(1, - 5)

Find the slope of

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

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 = - 8

Find the line with slope m=

- 8

and passing through

(1, - 5) : y = - 8 x + 3

y = - 8 x + 3

f (x) = 2 x e^{4 x}

n = 0 \sum \infty \frac{n ( n + 1 )}{2}

x \to - 1 - lim (\frac{x - 1}{x + 1})

\int_{0}^{\frac{1}{2}} cos (π x) d x

(2 x y + 3 y^{2}) d x - (x^{2} + 2 x y) d y = 0

What is the tangent of x^4-4x^3-2,\at x=1 ?
The tangent of x^4-4x^3-2,\at x=1 is y=-8x+3