What is the tangent of f(x)=x^3-6x^2+12x-9 ?

The tangent of f(x)=x^3-6x^2+12x-9 is y=(3a_{0}^2-12a_{0}+12)x-2a_{0}^3+6a_{0}^2-9

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tangent of f(x)=x^3-6x^2+12x-9

tangent of

f (x) = x^{3} - 6 x^{2} + 12 x - 9

y = (3 a_{0}^{2} - 12 a_{0} + 12) x - 2 a_{0}^{3} + 6 a_{0}^{2} - 9

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, a_{0}^{3} - 6 a_{0}^{2} + 12 a_{0} - 9)

Find the slope of

f (x) = x^{3} - 6 x^{2} + 12 x - 9 : \frac{df ( x )}{d x} = 3 x^{2} - 12 x + 12

Compute the slope of

f (x) = x^{3} - 6 x^{2} + 12 x - 9

x = a_{0} : m = 3 a_{0}^{2} - 12 a_{0} + 12

Find the line with slope m=

3 a_{0}^{2} - 12 a_{0} + 12

and passing through

(a_{0}, a_{0}^{3} - 6 a_{0}^{2} + 12 a_{0} - 9) : y = (3 a_{0}^{2} - 12 a_{0} + 12) x - 2 a_{0}^{3} + 6 a_{0}^{2} - 9

y = (3 a_{0}^{2} - 12 a_{0} + 12) x - 2 a_{0}^{3} + 6 a_{0}^{2} - 9

\frac{\partial}{\partial y} (- 0.2 y + 23)

\frac{\partial}{\partial x} (- 5 x y (9 - x - y))

\int \frac{1}{( 16 x ^{4} + 8 x ^{2} + 1 )} d x

\int_{0}^{2} 2 π (x + 1) (4 - 2 x) d x

\int_{0}^{1} (\frac{3}{1 + t ^{2}}) d t

What is the tangent of f(x)=x^3-6x^2+12x-9 ?
The tangent of f(x)=x^3-6x^2+12x-9 is y=(3a_{0}^2-12a_{0}+12)x-2a_{0}^3+6a_{0}^2-9