What is the tangent of f(x)=(3x+1)(6x-2) ?

The tangent of f(x)=(3x+1)(6x-2) is y=36a_{0}x-18a_{0}^2-2

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tangent of f(x)=(3x+1)(6x-2)

tangent of

f (x) = (3 x + 1) (6 x - 2)

y = 36 a_{0} x - 18 a_{0}^{2} - 2

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, (3 a_{0} + 1) (6 a_{0} - 2))

Find the slope of

f (x) = (3 x + 1) (6 x - 2) : \frac{df ( x )}{d x} = 36 x

Compute the slope of

f (x) = (3 x + 1) (6 x - 2)

x = a_{0} : m = 36 a_{0}

Find the line with slope m=

36 a_{0}

and passing through

(a_{0}, (3 a_{0} + 1) (6 a_{0} - 2)) : y = 36 a_{0} x - 18 a_{0}^{2} - 2

y = 36 a_{0} x - 18 a_{0}^{2} - 2

\frac{1}{s \cdot ( s ^{2} + 5 s + 4 )}

\frac{d}{d x} (- 6 cos (x))

\int \frac{x ^{2}}{( 4 - x ^{2} ) ^{\frac{1}{2}}} d x

\frac{d}{d x} (x^{2} + x^{4})

f (x) = \frac{1}{1 - x ^{2}}

What is the tangent of f(x)=(3x+1)(6x-2) ?
The tangent of f(x)=(3x+1)(6x-2) is y=36a_{0}x-18a_{0}^2-2