What is the tangent of f(x)=-2x^3,(2,-16) ?

The tangent of f(x)=-2x^3,(2,-16) is y=-24x+32

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tangent of f(x)=-2x^3,(2,-16)

tangent of

f (x) = - 2 x^{3}, (2, - 16)

y = - 24 x + 32

Solution steps

Find the slope of

f (x) = - 2 x^{3} : \frac{df ( x )}{d x} = - 6 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 = - 24

Find the line with slope m=

- 24

and passing through

(2, - 16) : y = - 24 x + 32

y = - 24 x + 32

y^{^{''}} + y^{^{'}} - 2 y = 4 sin (2 x)

\int e^{- 0.89 x} d x

\int \frac{d ^{2} y}{1} d x^{2}

\int - \frac{6 x}{5} d x

d er i v a t i v e (f (x))^{2}

What is the tangent of f(x)=-2x^3,(2,-16) ?
The tangent of f(x)=-2x^3,(2,-16) is y=-24x+32