What is the tangent of (x+1)(4x^2-4x-8) ?

The tangent of (x+1)(4x^2-4x-8) is y=(12a_{0}^2-12)x-8a_{0}^3-8

Popular >

tangent of (x+1)(4x^2-4x-8)

tangent of

(x + 1) (4 x^{2} - 4 x - 8)

y = (12 a_{0}^{2} - 12) x - 8 a_{0}^{3} - 8

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, (a_{0} + 1) (4 a_{0}^{2} - 4 a_{0} - 8))

Find the slope of

y = (x + 1) (4 x^{2} - 4 x - 8) : \frac{d y}{d x} = 12 x^{2} - 12

Compute the slope of

y = (x + 1) (4 x^{2} - 4 x - 8)

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

Find the line with slope m=

12 a_{0}^{2} - 12

and passing through

(a_{0}, (a_{0} + 1) (4 a_{0}^{2} - 4 a_{0} - 8)) : y = (12 a_{0}^{2} - 12) x - 8 a_{0}^{3} - 8

y = (12 a_{0}^{2} - 12) x - 8 a_{0}^{3} - 8

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

(- 12 x y^{2} - 12 y) + (- 12 x^{2} y - 12 x) y^{^{'}} = 0

\int e^{- 4 x} d x

\int \frac{( x ^{10} )}{x ^{2} + x - 2} d x

f (x) = \frac{4}{x}, at x = 3

What is the tangent of (x+1)(4x^2-4x-8) ?
The tangent of (x+1)(4x^2-4x-8) is y=(12a_{0}^2-12)x-8a_{0}^3-8