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

The tangent of f(x)=(x^3+1)(x^2-2),\at x=2 is y=60x-102

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

tangent of

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

y = 60 x - 102

Solution steps

Find the tangent point:

(2, 18)

Find the slope of

f (x) = (x^{3} + 1) (x^{2} - 2) : \frac{df ( x )}{d x} = 5 x^{4} - 6 x^{2} + 2 x

Compute the slope of

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

x = 2 : m = 60

Find the line with slope m=

60

and passing through

(2, 18) : y = 60 x - 102

y = 60 x - 102

f (x) = (x - 4) (3 - x)^{3}, at x = 5

\int x^{2} (3 x + x^{- 3}) d x

\int sin (- 3 x) d x

t \to π lim (- 9 sin (- \frac{t}{2}))

\int - 3 x + 4 x d x

What is the tangent of f(x)=(x^3+1)(x^2-2),\at x=2 ?
The tangent of f(x)=(x^3+1)(x^2-2),\at x=2 is y=60x-102