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

The tangent of f(x)=(x^2-12)(2x-9) is y=(6a_{0}^2-18a_{0}-24)x-4a_{0}^3+9a_{0}^2+108

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

tangent of

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

y = (6 a_{0}^{2} - 18 a_{0} - 24) x - 4 a_{0}^{3} + 9 a_{0}^{2} + 108

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

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

Find the slope of

f (x) = (x^{2} - 12) (2 x - 9) : \frac{df ( x )}{d x} = 6 x^{2} - 18 x - 24

Compute the slope of

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

x = a_{0} : m = 6 a_{0}^{2} - 18 a_{0} - 24

Find the line with slope m=

6 a_{0}^{2} - 18 a_{0} - 24

and passing through

(a_{0}, (a_{0}^{2} - 12) (2 a_{0} - 9)) : y = (6 a_{0}^{2} - 18 a_{0} - 24) x - 4 a_{0}^{3} + 9 a_{0}^{2} + 108

y = (6 a_{0}^{2} - 18 a_{0} - 24) x - 4 a_{0}^{3} + 9 a_{0}^{2} + 108

\int 3 (3 x + 12)^{4} d x

y^{^{'}} = 5 y

\frac{d}{d x} (arctan (\frac{x}{1 - x}))

z \to 0 lim ((2 z - 8)^{\frac{1}{3}})

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

What is the tangent of f(x)=(x^2-12)(2x-9) ?
The tangent of f(x)=(x^2-12)(2x-9) is y=(6a_{0}^2-18a_{0}-24)x-4a_{0}^3+9a_{0}^2+108