What is the tangent of g(x)=2-x^3 ?

The tangent of g(x)=2-x^3 is y=-3a_{0}^2x+2+2a_{0}^3

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tangent of g(x)=2-x^3

tangent of

g (x) = 2 - x^{3}

y = - 3 a_{0}^{2} x + 2 + 2 a_{0}^{3}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 2 - a_{0}^{3})

Find the slope of

g (x) = 2 - x^{3} : \frac{d g ( x )}{d x} = - 3 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 = - 3 a_{0}^{2}

Find the line with slope m=

- 3 a_{0}^{2}

and passing through

(a_{0}, 2 - a_{0}^{3}) : y = - 3 a_{0}^{2} x + 2 + 2 a_{0}^{3}

y = - 3 a_{0}^{2} x + 2 + 2 a_{0}^{3}

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

\int - t^{\frac{3}{2}} d t

10 e^{x}

\int_{2}^{3} \frac{x}{x ^{2} - 1} d x

\int \frac{ln ( x )}{1 + x ^{2}} d x

What is the tangent of g(x)=2-x^3 ?
The tangent of g(x)=2-x^3 is y=-3a_{0}^2x+2+2a_{0}^3