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

The tangent of f(x)=-3x^4,\at x=-1 is y=12x+9

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

tangent of

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

y = 12 x + 9

Solution steps

Find the tangent point:

(- 1, - 3)

Find the slope of

f (x) = - 3 x^{4} : \frac{df ( x )}{d x} = - 12 x^{3}

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 = 12

Find the line with slope m=

12

and passing through

(- 1, - 3) : y = 12 x + 9

y = 12 x + 9

\int_{- 2}^{3} (x^{2} + 4) - (- x^{2} - 1) d x

x \to \infty lim (8^{- x})

p (x) = 2 x^{4} + 3 x^{3} - 6 x^{2} + 9

\frac{d}{d x} (4 cos (3 x) - 3 sin (4 x))

\frac{d}{d x} (sin (x^{3} + 3))

What is the tangent of f(x)=-3x^4,\at x=-1 ?
The tangent of f(x)=-3x^4,\at x=-1 is y=12x+9