What is the tangent of f(x)=9+4x^2-2x^3,\at x=2 ?

The tangent of f(x)=9+4x^2-2x^3,\at x=2 is y=-8x+25

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tangent of f(x)=9+4x^2-2x^3,\at x=2

tangent of

f (x) = 9 + 4 x^{2} - 2 x^{3}, at x = 2

y = - 8 x + 25

Solution steps

Find the tangent point:

(2, 9)

Find the slope of

f (x) = 9 + 4 x^{2} - 2 x^{3} : \frac{df ( x )}{d x} = - 6 x^{2} + 8 x

Compute the slope of

f (x) = 9 + 4 x^{2} - 2 x^{3}

x = 2 : m = - 8

Find the line with slope m=

- 8

and passing through

(2, 9) : y = - 8 x + 25

y = - 8 x + 25

f (x) = 1 + 4 + x^{2}, at x = 0

f (x) = 8 x^{2}

f (2)

\frac{1}{2 9}

\int \frac{x ^{3}}{( x ^{4} + 1 )} d x

What is the tangent of f(x)=9+4x^2-2x^3,\at x=2 ?
The tangent of f(x)=9+4x^2-2x^3,\at x=2 is y=-8x+25