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

The tangent of f(x)=2x^3-4x+6 is y=(6a_{0}^2-4)x-4a_{0}^3+6

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tangent of f(x)=2x^3-4x+6

tangent of

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

y = (6 a_{0}^{2} - 4) x - 4 a_{0}^{3} + 6

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 2 a_{0}^{3} - 4 a_{0} + 6)

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 = a_{0} : m = 6 a_{0}^{2} - 4

Find the line with slope m=

6 a_{0}^{2} - 4

and passing through

(a_{0}, 2 a_{0}^{3} - 4 a_{0} + 6) : y = (6 a_{0}^{2} - 4) x - 4 a_{0}^{3} + 6

y = (6 a_{0}^{2} - 4) x - 4 a_{0}^{3} + 6

\int_{0}^{4} 0 d x

y^{^{'}} = 4 + e^{y - 4 x + 1}

\frac{400}{s ^{2} + 12 s + 400}

f (x) = x^{2} + 4, (- 3, 13)

\int \frac{cos ^{3} ( 3 t )}{sin ^{5} ( 3 t )} d t

What is the tangent of f(x)=2x^3-4x+6 ?
The tangent of f(x)=2x^3-4x+6 is y=(6a_{0}^2-4)x-4a_{0}^3+6