What is the tangent of x^3+4x^2+7x+1 ?

The tangent of x^3+4x^2+7x+1 is y=(3a_{0}^2+8a_{0}+7)x-2a_{0}^3-4a_{0}^2+1

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tangent of x^3+4x^2+7x+1

tangent of

x^{3} + 4 x^{2} + 7 x + 1

y = (3 a_{0}^{2} + 8 a_{0} + 7) x - 2 a_{0}^{3} - 4 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} + 4 a_{0}^{2} + 7 a_{0} + 1)

Find the slope of

y = x^{3} + 4 x^{2} + 7 x + 1 : \frac{d y}{d x} = 3 x^{2} + 8 x + 7

Compute the slope of

y = x^{3} + 4 x^{2} + 7 x + 1

x = a_{0} : m = 3 a_{0}^{2} + 8 a_{0} + 7

Find the line with slope m=

3 a_{0}^{2} + 8 a_{0} + 7

and passing through

(a_{0}, a_{0}^{3} + 4 a_{0}^{2} + 7 a_{0} + 1) : y = (3 a_{0}^{2} + 8 a_{0} + 7) x - 2 a_{0}^{3} - 4 a_{0}^{2} + 1

y = (3 a_{0}^{2} + 8 a_{0} + 7) x - 2 a_{0}^{3} - 4 a_{0}^{2} + 1

n = 2 \sum \infty \frac{6}{7 ^{n}}

x \to - 3 lim (\frac{2 x}{x ^{2} + 4})

x y^{^{'}} - 2 y = 2 x^{2}

y^{^{''}} + y^{^{'}} - 42 y = 0

\int \frac{5 x ^{2}}{( a ^{2} - x ^{2} ) ^{\frac{3}{2}}} d x

What is the tangent of x^3+4x^2+7x+1 ?
The tangent of x^3+4x^2+7x+1 is y=(3a_{0}^2+8a_{0}+7)x-2a_{0}^3-4a_{0}^2+1