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

The tangent of f(x)=2x^3-x^2+2,\at x=2 is y=20x-26

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

tangent of

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

y = 20 x - 26

Solution steps

Find the tangent point:

(2, 14)

Find the slope of

f (x) = 2 x^{3} - x^{2} + 2 : \frac{df ( x )}{d x} = 6 x^{2} - 2 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 = 20

Find the line with slope m=

20

and passing through

(2, 14) : y = 20 x - 26

y = 20 x - 26

\frac{\partial}{\partial y} (e^{- y} (x^{2} - y^{2}))

\frac{d}{d x} (\frac{1}{14} (e^{7 x} + e^{- 7 x}))

\int_{0}^{1} π (tan (\frac{π}{4} y))^{2} d y

\frac{\partial}{\partial x} (\frac{5 x + 4 y}{3 x + 2 y})

\int (tan (x))^{2} sec (x) d x

What is the tangent of f(x)=2x^3-x^2+2,\at x=2 ?
The tangent of f(x)=2x^3-x^2+2,\at x=2 is y=20x-26