What is the tangent of f(x)=x^3+2x^2+4 ?

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

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tangent of f(x)=x^3+2x^2+4

tangent of

f (x) = x^{3} + 2 x^{2} + 4

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

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

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

Find the slope of

f (x) = x^{3} + 2 x^{2} + 4 : \frac{df ( x )}{d x} = 3 x^{2} + 4 x

Compute the slope of

f (x) = x^{3} + 2 x^{2} + 4

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

Find the line with slope m=

3 a_{0}^{2} + 4 a_{0}

and passing through

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

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

\frac{d}{d x} (18)

f (x) = \frac{2 - x cos ( π x )}{x - 2}

y = 3 x, y = \frac{3}{5} x, y = 28 - x^{2}

\frac{\partial}{\partial x} (x^{5} e^{3 x y})

\frac{\partial}{\partial y} (2 + y e^{x y})

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