What is the tangent of f(x)=2x^3+36x^2+192x+8 ?

The tangent of f(x)=2x^3+36x^2+192x+8 is y=(6a_{0}^2+72a_{0}+192)x-4a_{0}^3-36a_{0}^2+8

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tangent of f(x)=2x^3+36x^2+192x+8

tangent of

f (x) = 2 x^{3} + 36 x^{2} + 192 x + 8

y = (6 a_{0}^{2} + 72 a_{0} + 192) x - 4 a_{0}^{3} - 36 a_{0}^{2} + 8

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 2 a_{0}^{3} + 36 a_{0}^{2} + 192 a_{0} + 8)

Find the slope of

f (x) = 2 x^{3} + 36 x^{2} + 192 x + 8 : \frac{df ( x )}{d x} = 6 x^{2} + 72 x + 192

Compute the slope of

f (x) = 2 x^{3} + 36 x^{2} + 192 x + 8

x = a_{0} : m = 6 a_{0}^{2} + 72 a_{0} + 192

Find the line with slope m=

6 a_{0}^{2} + 72 a_{0} + 192

and passing through

(a_{0}, 2 a_{0}^{3} + 36 a_{0}^{2} + 192 a_{0} + 8) : y = (6 a_{0}^{2} + 72 a_{0} + 192) x - 4 a_{0}^{3} - 36 a_{0}^{2} + 8

y = (6 a_{0}^{2} + 72 a_{0} + 192) x - 4 a_{0}^{3} - 36 a_{0}^{2} + 8

x \to 0 + lim (\frac{e ^{x}}{1 - e ^{- x}})

\frac{d}{d x} (4 e^{6 x})

f (x) = sin (3 x + 1)

\frac{d}{d x} ((4 x)^{6 x})

x \to π lim (x cos (\frac{7 x}{6}))

What is the tangent of f(x)=2x^3+36x^2+192x+8 ?
The tangent of f(x)=2x^3+36x^2+192x+8 is y=(6a_{0}^2+72a_{0}+192)x-4a_{0}^3-36a_{0}^2+8