What is the tangent of f(x)= 1/5 x^{10}-1/3 x^6-x+9 ?

The tangent of f(x)= 1/5 x^{10}-1/3 x^6-x+9 is y=(2a_{0}^9-2a_{0}^5-1)x-9/5 a_{0}^{10}+5/3 a_{0}^6+9

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tangent of f(x)= 1/5 x^{10}-1/3 x^6-x+9

tangent of

f (x) = \frac{1}{5} x^{10} - \frac{1}{3} x^{6} - x + 9

y = (2 a_{0}^{9} - 2 a_{0}^{5} - 1) x - \frac{9}{5} a_{0}^{10} + \frac{5}{3} a_{0}^{6} + 9

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, \frac{1}{5} a_{0}^{10} - \frac{1}{3} a_{0}^{6} - a_{0} + 9)

Find the slope of

f (x) = \frac{1}{5} x^{10} - \frac{1}{3} x^{6} - x + 9 : \frac{df ( x )}{d x} = 2 x^{9} - 2 x^{5} - 1

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = 2 a_{0}^{9} - 2 a_{0}^{5} - 1

Find the line with slope m=

2 a_{0}^{9} - 2 a_{0}^{5} - 1

and passing through

(a_{0}, \frac{1}{5} a_{0}^{10} - \frac{1}{3} a_{0}^{6} - a_{0} + 9) : y = (2 a_{0}^{9} - 2 a_{0}^{5} - 1) x - \frac{9}{5} a_{0}^{10} + \frac{5}{3} a_{0}^{6} + 9

y = (2 a_{0}^{9} - 2 a_{0}^{5} - 1) x - \frac{9}{5} a_{0}^{10} + \frac{5}{3} a_{0}^{6} + 9

What is the tangent of f(x)= 1/5 x^{10}-1/3 x^6-x+9 ?
The tangent of f(x)= 1/5 x^{10}-1/3 x^6-x+9 is y=(2a_{0}^9-2a_{0}^5-1)x-9/5 a_{0}^{10}+5/3 a_{0}^6+9