What is the tangent of x^4-3x^3+2x^2-x+1 ?

The tangent of x^4-3x^3+2x^2-x+1 is y=(4a_{0}^3-9a_{0}^2+4a_{0}-1)x-3a_{0}^4+6a_{0}^3-2a_{0}^2+1

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tangent of x^4-3x^3+2x^2-x+1

tangent of

x^{4} - 3 x^{3} + 2 x^{2} - x + 1

y = (4 a_{0}^{3} - 9 a_{0}^{2} + 4 a_{0} - 1) x - 3 a_{0}^{4} + 6 a_{0}^{3} - 2 a_{0}^{2} + 1

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, a_{0}^{4} - 3 a_{0}^{3} + 2 a_{0}^{2} - a_{0} + 1)

Find the slope of

y = x^{4} - 3 x^{3} + 2 x^{2} - x + 1 : \frac{d y}{d x} = 4 x^{3} - 9 x^{2} + 4 x - 1

Compute the slope of

y = x^{4} - 3 x^{3} + 2 x^{2} - x + 1

x = a_{0} : m = 4 a_{0}^{3} - 9 a_{0}^{2} + 4 a_{0} - 1

Find the line with slope m=

4 a_{0}^{3} - 9 a_{0}^{2} + 4 a_{0} - 1

and passing through

(a_{0}, a_{0}^{4} - 3 a_{0}^{3} + 2 a_{0}^{2} - a_{0} + 1) : y = (4 a_{0}^{3} - 9 a_{0}^{2} + 4 a_{0} - 1) x - 3 a_{0}^{4} + 6 a_{0}^{3} - 2 a_{0}^{2} + 1

y = (4 a_{0}^{3} - 9 a_{0}^{2} + 4 a_{0} - 1) x - 3 a_{0}^{4} + 6 a_{0}^{3} - 2 a_{0}^{2} + 1

f (x) = x^{5} cos (2 x)

\int_{- 2}^{3} (x^{3} - 6 x) d x

\int \frac{x}{x ^{2} sin ^{2} ( ln ( x ) )} d x

\frac{( x + 5 )}{( x - 5 )}

f (x) = x^{2} - tan (x), at x = 0

What is the tangent of x^4-3x^3+2x^2-x+1 ?
The tangent of x^4-3x^3+2x^2-x+1 is y=(4a_{0}^3-9a_{0}^2+4a_{0}-1)x-3a_{0}^4+6a_{0}^3-2a_{0}^2+1