What is the tangent of f(x)=x^3+x+5,\at x=-2 ?

The tangent of f(x)=x^3+x+5,\at x=-2 is y=13x+21

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tangent of f(x)=x^3+x+5,\at x=-2

tangent of

f (x) = x^{3} + x + 5, at x = - 2

y = 13 x + 21

Solution steps

Find the tangent point:

(- 2, - 5)

Find the slope of

f (x) = x^{3} + x + 5 : \frac{df ( x )}{d x} = 3 x^{2} + 1

Compute the slope of

f (x) = x^{3} + x + 5

x = - 2 : m = 13

Find the line with slope m=

13

and passing through

(- 2, - 5) : y = 13 x + 21

y = 13 x + 21

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

\frac{d}{d x} (arctan (\frac{x + y}{1 - x y}))

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

\frac{\partial}{\partial y} (3 x^{2} (5 x + 7 y))

f (x) = x^{3} - 6 x^{2} + 9 x

What is the tangent of f(x)=x^3+x+5,\at x=-2 ?
The tangent of f(x)=x^3+x+5,\at x=-2 is y=13x+21