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

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

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

tangent of

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

y = 3 x - 1

Solution steps

Find the tangent point:

(1, 2)

Find the slope of

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

Compute the slope of

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

x = 1 : m = 3

Find the line with slope m=

3

and passing through

(1, 2) : y = 3 x - 1

y = 3 x - 1

\frac{d}{d x} (- 3 sin (x) - 2)

\frac{d}{d x} (x - 2 ln (x))

f (x) = 11 + 9 x - x^{3}, at x = 1

\frac{\partial}{\partial x} (\frac{1}{( x - y ) ^{2}})

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