What is the tangent of f(x)=2x^3+6x^2-49x+24 ?

The tangent of f(x)=2x^3+6x^2-49x+24 is y=(6a_{0}^2+12a_{0}-49)x-4a_{0}^3-6a_{0}^2+24

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tangent of f(x)=2x^3+6x^2-49x+24

tangent of

f (x) = 2 x^{3} + 6 x^{2} - 49 x + 24

y = (6 a_{0}^{2} + 12 a_{0} - 49) x - 4 a_{0}^{3} - 6 a_{0}^{2} + 24

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 2 a_{0}^{3} + 6 a_{0}^{2} - 49 a_{0} + 24)

Find the slope of

f (x) = 2 x^{3} + 6 x^{2} - 49 x + 24 : \frac{df ( x )}{d x} = 6 x^{2} + 12 x - 49

Compute the slope of

f (x) = 2 x^{3} + 6 x^{2} - 49 x + 24

x = a_{0} : m = 6 a_{0}^{2} + 12 a_{0} - 49

Find the line with slope m=

6 a_{0}^{2} + 12 a_{0} - 49

and passing through

(a_{0}, 2 a_{0}^{3} + 6 a_{0}^{2} - 49 a_{0} + 24) : y = (6 a_{0}^{2} + 12 a_{0} - 49) x - 4 a_{0}^{3} - 6 a_{0}^{2} + 24

y = (6 a_{0}^{2} + 12 a_{0} - 49) x - 4 a_{0}^{3} - 6 a_{0}^{2} + 24

\frac{\partial}{\partial x} (z arctan (\frac{3 y}{x}))

\int_{- 2.5}^{0} e^{3 x} d x

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

\frac{\partial}{\partial x} (2 x y + x^{3})

\frac{d}{d x} (1 - x^{4})

What is the tangent of f(x)=2x^3+6x^2-49x+24 ?
The tangent of f(x)=2x^3+6x^2-49x+24 is y=(6a_{0}^2+12a_{0}-49)x-4a_{0}^3-6a_{0}^2+24