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

The tangent of f(x)=9x^2-x^3,\at x=2 is y=24x-20

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

tangent of

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

y = 24 x - 20

Solution steps

Find the tangent point:

(2, 28)

Find the slope of

f (x) = 9 x^{2} - x^{3} : \frac{df ( x )}{d x} = 18 x - 3 x^{2}

Compute the slope of

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

x = 2 : m = 24

Find the line with slope m=

24

and passing through

(2, 28) : y = 24 x - 20

y = 24 x - 20

\int e^{2} sin (x) d x

\int_{0}^{3} 3 x - 1 d x

y^{^{''}} - 9 y^{^{'}} + 2 y = x e^{x}

\int \frac{x + 2}{4 x - x ^{2}} d x

\frac{d}{d x} (\frac{2}{( x + 1 ) ^{3}})

What is the tangent of f(x)=9x^2-x^3,\at x=2 ?
The tangent of f(x)=9x^2-x^3,\at x=2 is y=24x-20