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

The tangent of f(x)=4+x-2x^2-3x^3,\at x=2 is y=-43x+60

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tangent of f(x)=4+x-2x^2-3x^3,\at x=2

tangent of

f (x) = 4 + x - 2 x^{2} - 3 x^{3}, at x = 2

y = - 43 x + 60

Solution steps

Find the tangent point:

(2, - 26)

Find the slope of

f (x) = 4 + x - 2 x^{2} - 3 x^{3} : \frac{df ( x )}{d x} = - 9 x^{2} - 4 x + 1

Compute the slope of

f (x) = 4 + x - 2 x^{2} - 3 x^{3}

x = 2 : m = - 43

Find the line with slope m=

- 43

and passing through

(2, - 26) : y = - 43 x + 60

y = - 43 x + 60

x \to 1 lim (\frac{2}{x ^{3} - 1})

\frac{d}{d x} (sin (x^{c o s (x)}))

y^{^{'}} = 3 + e^{- 2 t}

t^{2} sin (4 t)

\frac{d y}{d t} + e^{t} y = 12 e^{t}

What is the tangent of f(x)=4+x-2x^2-3x^3,\at x=2 ?
The tangent of f(x)=4+x-2x^2-3x^3,\at x=2 is y=-43x+60