What is the tangent of g(x)=-x^5+4x^2-2x+2 ?

The tangent of g(x)=-x^5+4x^2-2x+2 is y=(-5a_{0}^4+8a_{0}-2)x+4a_{0}^5-4a_{0}^2+2

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tangent of g(x)=-x^5+4x^2-2x+2

tangent of

g (x) = - x^{5} + 4 x^{2} - 2 x + 2

y = (- 5 a_{0}^{4} + 8 a_{0} - 2) x + 4 a_{0}^{5} - 4 a_{0}^{2} + 2

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, - a_{0}^{5} + 4 a_{0}^{2} - 2 a_{0} + 2)

Find the slope of

g (x) = - x^{5} + 4 x^{2} - 2 x + 2 : \frac{d g ( x )}{d x} = - 5 x^{4} + 8 x - 2

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 = - 5 a_{0}^{4} + 8 a_{0} - 2

Find the line with slope m=

- 5 a_{0}^{4} + 8 a_{0} - 2

and passing through

(a_{0}, - a_{0}^{5} + 4 a_{0}^{2} - 2 a_{0} + 2) : y = (- 5 a_{0}^{4} + 8 a_{0} - 2) x + 4 a_{0}^{5} - 4 a_{0}^{2} + 2

y = (- 5 a_{0}^{4} + 8 a_{0} - 2) x + 4 a_{0}^{5} - 4 a_{0}^{2} + 2

\int \frac{sin ( 2 x )}{18 + cos ^{2} ( x )} d x

\frac{d}{d x} (\frac{4 x ^{5} - 5 x ^{4} + 6 x - 2}{x ^{4}})

y^{^{'}} + 3 y + 5 = 0

\frac{d}{d x} (3 x^{2} + 12 x + 6)

Y = (x^{\frac{1}{2}}), Y = 6 - x, Y = 0

What is the tangent of g(x)=-x^5+4x^2-2x+2 ?
The tangent of g(x)=-x^5+4x^2-2x+2 is y=(-5a_{0}^4+8a_{0}-2)x+4a_{0}^5-4a_{0}^2+2