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

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

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tangent of f(x)=2x^3-4x+6,\at x=1

tangent of

f (x) = 2 x^{3} - 4 x + 6, at x = 1

y = 2 x + 2

Solution steps

Find the tangent point:

(1, 4)

Find the slope of

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

Compute the slope of

f (x) = 2 x^{3} - 4 x + 6

x = 1 : m = 2

Find the line with slope m=

2

and passing through

(1, 4) : y = 2 x + 2

y = 2 x + 2

\int_{2}^{7} g (x) d x

\frac{d}{d x} (\frac{x ^{2} - 1}{3 x})

\int e^{- a t} d t

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

x \to \infty lim (1.3 x)

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