What is the tangent of (x^3)/6-3x^2+((77x))/6-8,\at x=5 ?

The tangent of (x^3)/6-3x^2+((77x))/6-8,\at x=5 is y=-14/3 x+76/3

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tangent of (x^3)/6-3x^2+((77x))/6-8,\at x=5

tangent of

\frac{x ^{3}}{6} - 3 x^{2} + \frac{( 77 x )}{6} - 8, at x = 5

y = - \frac{14}{3} x + \frac{76}{3}

Solution steps

Find the tangent point:

(5, 2)

Find the slope of

y = \frac{x ^{3}}{6} - 3 x^{2} + \frac{77 x}{6} - 8 : \frac{d y}{d x} = - 6 x + \frac{1}{6} (3 x^{2} + 77)

Compute the slope of

y = \frac{x ^{3}}{6} - 3 x^{2} + \frac{77 x}{6} - 8

x = 5 : m = - \frac{14}{3}

Find the line with slope m=

- \frac{14}{3}

and passing through

(5, 2) : y = - \frac{14}{3} x + \frac{76}{3}

y = - \frac{14}{3} x + \frac{76}{3}

t^{3} ln (t)

\frac{d y}{d x} = \frac{y}{2 + x ^{2}}

x \to 0 lim (\frac{e ^{x} - 2}{x})

x \to - 8 lim (\frac{8 + x}{8 + x})

\int (x^{4} + x^{3})^{12} (4 x^{3} + 3 x^{2}) d x

What is the tangent of (x^3)/6-3x^2+((77x))/6-8,\at x=5 ?
The tangent of (x^3)/6-3x^2+((77x))/6-8,\at x=5 is y=-14/3 x+76/3