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

The tangent of f(x)=-x^3+4x^2,\at x=3 is y=-3x+18

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

tangent of

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

y = - 3 x + 18

Solution steps

Find the tangent point:

(3, 9)

Find the slope of

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

Compute the slope of

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

x = 3 : m = - 3

Find the line with slope m=

- 3

and passing through

(3, 9) : y = - 3 x + 18

y = - 3 x + 18

y = x^{3} - 2 x^{2} + 3, y = 3 x^{2} + 5 x - 22

y = x^{\frac{15}{14}}, y = 13 x^{\frac{1}{14}}

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

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

What is the tangent of f(x)=-x^3+4x^2,\at x=3 ?
The tangent of f(x)=-x^3+4x^2,\at x=3 is y=-3x+18