What is the tangent of f(x)=5+5x^2-2x^3,\at x=2 ?

The tangent of f(x)=5+5x^2-2x^3,\at x=2 is y=-4x+17

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tangent of f(x)=5+5x^2-2x^3,\at x=2

tangent of

f (x) = 5 + 5 x^{2} - 2 x^{3}, at x = 2

y = - 4 x + 17

Solution steps

Find the tangent point:

(2, 9)

Find the slope of

f (x) = 5 + 5 x^{2} - 2 x^{3} : \frac{df ( x )}{d x} = - 6 x^{2} + 10 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 = - 4

Find the line with slope m=

- 4

and passing through

(2, 9) : y = - 4 x + 17

y = - 4 x + 17

\frac{\partial}{\partial x} (\frac{G ( m , x ) m}{x ^{2}})

\int \frac{- 3 e ^{5 x}}{5 x} d x

e^{l n (c o s (x^{2}))} 2 x + 1

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

\int 3 sin (t) cos^{2} (t) d t

What is the tangent of f(x)=5+5x^2-2x^3,\at x=2 ?
The tangent of f(x)=5+5x^2-2x^3,\at x=2 is y=-4x+17