What is the tangent of f(x)=2(x^2-1)^3,\at x=2 ?

The tangent of f(x)=2(x^2-1)^3,\at x=2 is y=216x-378

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tangent of f(x)=2(x^2-1)^3,\at x=2

tangent of

f (x) = 2 (x^{2} - 1)^{3}, at x = 2

y = 216 x - 378

Solution steps

Find the tangent point:

(2, 54)

Find the slope of

f (x) = 2 (x^{2} - 1)^{3} : \frac{df ( x )}{d x} = 12 x (x^{2} - 1)^{2}

Compute the slope of

f (x) = 2 (x^{2} - 1)^{3}

x = 2 : m = 216

Find the line with slope m=

216

and passing through

(2, 54) : y = 216 x - 378

y = 216 x - 378

x \to - 8 + lim (\frac{2 x}{x + 8})

f (x) = \frac{2}{4 x + 3}

f (x) = \frac{1 - 3 x}{1 + 3 x}

x \to \frac{3}{4} lim (\frac{4 x ^{2} - 9}{2 x - 3})

\int_{0}^{3} 3 - x d x

What is the tangent of f(x)=2(x^2-1)^3,\at x=2 ?
The tangent of f(x)=2(x^2-1)^3,\at x=2 is y=216x-378