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

The tangent of f(x)=(x^3+2)(3x^2-4x+2),(1,3) is y=9x-6

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

tangent of

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

y = 9 x - 6

Solution steps

Find the slope of

f (x) = (x^{3} + 2) (3 x^{2} - 4 x + 2) : \frac{df ( x )}{d x} = 15 x^{4} - 16 x^{3} + 6 x^{2} + 12 x - 8

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 = 9

Find the line with slope m=

9

and passing through

(1, 3) : y = 9 x - 6

y = 9 x - 6

\int \frac{3 x ^{2} + 3 x + 1}{x ^{3} + 2 x ^{2} + 2 x + 1} d x

(1, 2), (5, - 3)

y = \frac{1}{x + 1}

\int \frac{7 x ^{2}}{x ^{2} - 4} d x

\int x^{2} + x y^{2} + z d x

What is the tangent of f(x)=(x^3+2)(3x^2-4x+2),(1,3) ?
The tangent of f(x)=(x^3+2)(3x^2-4x+2),(1,3) is y=9x-6