What is the tangent of f(x)=(x^2-18)(2x-3) ?

The tangent of f(x)=(x^2-18)(2x-3) is y=(6a_{0}^2-6a_{0}-36)x-4a_{0}^3+3a_{0}^2+54

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tangent of f(x)=(x^2-18)(2x-3)

tangent of

f (x) = (x^{2} - 18) (2 x - 3)

y = (6 a_{0}^{2} - 6 a_{0} - 36) x - 4 a_{0}^{3} + 3 a_{0}^{2} + 54

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, (a_{0}^{2} - 18) (2 a_{0} - 3))

Find the slope of

f (x) = (x^{2} - 18) (2 x - 3) : \frac{df ( x )}{d x} = 6 x^{2} - 6 x - 36

Compute the slope of

f (x) = (x^{2} - 18) (2 x - 3)

x = a_{0} : m = 6 a_{0}^{2} - 6 a_{0} - 36

Find the line with slope m=

6 a_{0}^{2} - 6 a_{0} - 36

and passing through

(a_{0}, (a_{0}^{2} - 18) (2 a_{0} - 3)) : y = (6 a_{0}^{2} - 6 a_{0} - 36) x - 4 a_{0}^{3} + 3 a_{0}^{2} + 54

y = (6 a_{0}^{2} - 6 a_{0} - 36) x - 4 a_{0}^{3} + 3 a_{0}^{2} + 54

\frac{d}{d x} (\frac{- x ^{3}}{y ^{3}})

\int t e^{a t} d t

n = 0 \sum \infty \frac{8}{n} - \frac{8}{n + 1}

5^{^{'}}

\int x e^{\frac{1}{2} x^{2}} d x

What is the tangent of f(x)=(x^2-18)(2x-3) ?
The tangent of f(x)=(x^2-18)(2x-3) is y=(6a_{0}^2-6a_{0}-36)x-4a_{0}^3+3a_{0}^2+54