What is the tangent of 1/3 x^3-3x^2+10x+6,\at x=2 ?

The tangent of 1/3 x^3-3x^2+10x+6,\at x=2 is y=2x+38/3

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tangent of 1/3 x^3-3x^2+10x+6,\at x=2

tangent of

\frac{1}{3} x^{3} - 3 x^{2} + 10 x + 6, at x = 2

y = 2 x + \frac{38}{3}

Solution steps

Find the tangent point:

(2, \frac{50}{3})

Find the slope of

y = \frac{1}{3} x^{3} - 3 x^{2} + 10 x + 6 : \frac{d y}{d x} = x^{2} - 6 x + 10

Compute the slope of

y = \frac{1}{3} x^{3} - 3 x^{2} + 10 x + 6

x = 2 : m = 2

Find the line with slope m=

2

and passing through

(2, \frac{50}{3}) : y = 2 x + \frac{38}{3}

y = 2 x + \frac{38}{3}

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

y = x sin (\frac{6}{x})

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

\frac{d}{d x} (e^{0.1 x^{2}})

What is the tangent of 1/3 x^3-3x^2+10x+6,\at x=2 ?
The tangent of 1/3 x^3-3x^2+10x+6,\at x=2 is y=2x+38/3