What is the tangent of f(x)=8+4x^2-2x^3,\at x=2 ?

The tangent of f(x)=8+4x^2-2x^3,\at x=2 is y=-8x+24

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tangent of f(x)=8+4x^2-2x^3,\at x=2

tangent of

f (x) = 8 + 4 x^{2} - 2 x^{3}, at x = 2

y = - 8 x + 24

Solution steps

Find the tangent point:

(2, 8)

Find the slope of

f (x) = 8 + 4 x^{2} - 2 x^{3} : \frac{df ( x )}{d x} = - 6 x^{2} + 8 x

Compute the slope of

f (x) = 8 + 4 x^{2} - 2 x^{3}

x = 2 : m = - 8

Find the line with slope m=

- 8

and passing through

(2, 8) : y = - 8 x + 24

y = - 8 x + 24

x^{3} y^{^{'}} = x^{2} y - y^{3}

\int cos^{3} (\frac{x}{13}) d x

x \to 0 lim (x + 6)

y = (9 x^{4} + 5 x)^{3}

\int e^{8 x + 7} d x

What is the tangent of f(x)=8+4x^2-2x^3,\at x=2 ?
The tangent of f(x)=8+4x^2-2x^3,\at x=2 is y=-8x+24