What is the tangent of x^3-25x+5 ?

The tangent of x^3-25x+5 is y=(3a_{0}^2-25)x-2a_{0}^3+5

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tangent of x^3-25x+5

tangent of

x^{3} - 25 x + 5

y = (3 a_{0}^{2} - 25) x - 2 a_{0}^{3} + 5

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, a_{0}^{3} - 25 a_{0} + 5)

Find the slope of

y = x^{3} - 25 x + 5 : \frac{d y}{d x} = 3 x^{2} - 25

Compute the slope of

y = x^{3} - 25 x + 5

x = a_{0} : m = 3 a_{0}^{2} - 25

Find the line with slope m=

3 a_{0}^{2} - 25

and passing through

(a_{0}, a_{0}^{3} - 25 a_{0} + 5) : y = (3 a_{0}^{2} - 25) x - 2 a_{0}^{3} + 5

y = (3 a_{0}^{2} - 25) x - 2 a_{0}^{3} + 5

\frac{d}{d x} (2 x^{3} - 11.7 x^{2} + 17.7 x - 5)

(1 + x)^{\frac{1}{2}}

\frac{d}{d x} ((x^{\frac{1}{2}} + y^{\frac{1}{2}})^{2})

(2 x^{3} - y) d x - (x + 2 y) d y = 0

y^{^{'}} = \frac{y}{t} + 2 t + 1

What is the tangent of x^3-25x+5 ?
The tangent of x^3-25x+5 is y=(3a_{0}^2-25)x-2a_{0}^3+5