What is the tangent of f(x)=8+4x^2-2x^3,\at x=1 ?

The tangent of f(x)=8+4x^2-2x^3,\at x=1 is y=2x+8

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tangent of f(x)=8+4x^2-2x^3,\at x=1

tangent of

f (x) = 8 + 4 x^{2} - 2 x^{3}, at x = 1

y = 2 x + 8

Solution steps

Find the tangent point:

(1, 10)

Find the slope of

f (x) = 8 + 4 x^{2} - 2 x^{3} : \frac{df ( x )}{d x} = - 6 x^{2} + 8 x

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 = 2

Find the line with slope m=

2

and passing through

(1, 10) : y = 2 x + 8

y = 2 x + 8

What is the tangent of f(x)=8+4x^2-2x^3,\at x=1 ?
The tangent of f(x)=8+4x^2-2x^3,\at x=1 is y=2x+8