What is the tangent of f(x)=9+4x^2-2x^3,\at x=1 ?

The tangent of f(x)=9+4x^2-2x^3,\at x=1 is y=2x+9

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tangent of f(x)=9+4x^2-2x^3,\at x=1

tangent of

f (x) = 9 + 4 x^{2} - 2 x^{3}, at x = 1

y = 2 x + 9

Solution steps

Find the tangent point:

(1, 11)

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 = 1 : m = 2

Find the line with slope m=

2

and passing through

(1, 11) : y = 2 x + 9

y = 2 x + 9

y = ln (ln (x^{3}))

y = \frac{sin ( x )}{1 + sin ( x )}

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

\int \frac{x ^{3}}{x ^{2} - 49} d x

\frac{d}{d x} (1 - e^{- x^{2}})

What is the tangent of f(x)=9+4x^2-2x^3,\at x=1 ?
The tangent of f(x)=9+4x^2-2x^3,\at x=1 is y=2x+9