What is the tangent of 1/3 x^3-7x+5 ?

The tangent of 1/3 x^3-7x+5 is y=(a_{0}^2-7)x-2/3 a_{0}^3+5

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tangent of 1/3 x^3-7x+5

tangent of

\frac{1}{3} x^{3} - 7 x + 5

y = (a_{0}^{2} - 7) x - \frac{2}{3} a_{0}^{3} + 5

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, \frac{1}{3} a_{0}^{3} - 7 a_{0} + 5)

Find the slope of

y = \frac{1}{3} x^{3} - 7 x + 5 : \frac{d y}{d x} = x^{2} - 7

Compute the slope of

y = \frac{1}{3} x^{3} - 7 x + 5

x = a_{0} : m = a_{0}^{2} - 7

Find the line with slope m=

a_{0}^{2} - 7

and passing through

(a_{0}, \frac{1}{3} a_{0}^{3} - 7 a_{0} + 5) : y = (a_{0}^{2} - 7) x - \frac{2}{3} a_{0}^{3} + 5

y = (a_{0}^{2} - 7) x - \frac{2}{3} a_{0}^{3} + 5

n = 1 \sum \infty \frac{x ^{n}}{2 ^{n}}

x \to 0 lim (\frac{f ( x )}{g ( x )})

\int \frac{a + b x ^{3}}{4 a x + b x ^{4}} d x

\frac{d}{d z} (\frac{x y}{z})

\int (3 sin (t) cos^{2} (t)) d t

What is the tangent of 1/3 x^3-7x+5 ?
The tangent of 1/3 x^3-7x+5 is y=(a_{0}^2-7)x-2/3 a_{0}^3+5