What is the tangent of y=2x^3-x^2+2(1.3) ?

The tangent of y=2x^3-x^2+2(1.3) is y=(6a_{0}^2-2a_{0})x-4a_{0}^3+a_{0}^2+2.6

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tangent of y=2x^3-x^2+2(1.3)

tangent of

y = 2 x^{3} - x^{2} + 2 (1.3)

y = (6 a_{0}^{2} - 2 a_{0}) x - 4 a_{0}^{3} + a_{0}^{2} + 2.6

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 2 a_{0}^{3} - a_{0}^{2} + 2.6)

Find the slope of

y = 2 x^{3} - x^{2} + 2 \cdot 1.3 : \frac{d y}{d x} = 6 x^{2} - 2 x

Compute the slope of

y = 2 x^{3} - x^{2} + 2 \cdot 1.3

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

Find the line with slope m=

6 a_{0}^{2} - 2 a_{0}

and passing through

(a_{0}, 2 a_{0}^{3} - a_{0}^{2} + 2.6) : y = (6 a_{0}^{2} - 2 a_{0}) x - 4 a_{0}^{3} + a_{0}^{2} + 2.6

y = (6 a_{0}^{2} - 2 a_{0}) x - 4 a_{0}^{3} + a_{0}^{2} + 2.6

(10.4) (4.15)

\int \frac{e ^{x}}{1 + e ^{x}} d x

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

y = \frac{x ^{2}}{x - 4}

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

What is the tangent of y=2x^3-x^2+2(1.3) ?
The tangent of y=2x^3-x^2+2(1.3) is y=(6a_{0}^2-2a_{0})x-4a_{0}^3+a_{0}^2+2.6