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

The tangent of f(x)=(2x^3+1)^2,(-1,1) is y=-12x-11

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

tangent of

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

y = - 12 x - 11

Solution steps

Find the slope of

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

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = - 12

Find the line with slope m=

- 12

and passing through

(- 1, 1) : y = - 12 x - 11

y = - 12 x - 11

x \to 0 lim (\frac{( e e ^{- a x} - e ^{- b x} )}{x})

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

x = 36 - y^{2}, x = y^{2} - 36

\int t - cos (t) d t

\frac{d}{d x} (\frac{3 x ^{2} - 18 x + 25}{x ^{2} - 6 x + 8})

What is the tangent of f(x)=(2x^3+1)^2,(-1,1) ?
The tangent of f(x)=(2x^3+1)^2,(-1,1) is y=-12x-11