What is the tangent of x^3+x^2+5x-1 ?

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

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tangent of x^3+x^2+5x-1

tangent of

x^{3} + x^{2} + 5 x - 1

y = (3 a_{0}^{2} + 2 a_{0} + 5) x - 2 a_{0}^{3} - 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}^{3} + a_{0}^{2} + 5 a_{0} - 1)

Find the slope of

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

Compute the slope of

y = x^{3} + x^{2} + 5 x - 1

x = a_{0} : m = 3 a_{0}^{2} + 2 a_{0} + 5

Find the line with slope m=

3 a_{0}^{2} + 2 a_{0} + 5

and passing through

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

y = (3 a_{0}^{2} + 2 a_{0} + 5) x - 2 a_{0}^{3} - a_{0}^{2} - 1

\int 3^{2 x} e^{x} d x

x \to - 4 lim (∣ x ∣)

\frac{d}{d x} (ln (x^{3} + 4))

h (x) = ln (\frac{x ^{2} - 1}{x ^{2} + 1})

y^{^{'}} = 2 cos (5 x)

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