What is the tangent of f(x)=-x^3-4x-3,\at x=-1 ?

The tangent of f(x)=-x^3-4x-3,\at x=-1 is y=-7x-5

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tangent of f(x)=-x^3-4x-3,\at x=-1

tangent of

f (x) = - x^{3} - 4 x - 3, at x = - 1

y = - 7 x - 5

Solution steps

Find the tangent point:

(- 1, 2)

Find the slope of

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

Compute the slope of

f (x) = - x^{3} - 4 x - 3

x = - 1 : m = - 7

Find the line with slope m=

- 7

and passing through

(- 1, 2) : y = - 7 x - 5

y = - 7 x - 5

\frac{1}{( s - 2 ) ( s - 2 )}

(x^{2} + x) e^{x}

x \to 1 + lim (\frac{1}{1 - x})

8 x^{2} y^{^{'}} = y^{^{'}} + 6 x e^{- y}

\int \frac{e ^{2 y} - y}{e ^{y}} d y

What is the tangent of f(x)=-x^3-4x-3,\at x=-1 ?
The tangent of f(x)=-x^3-4x-3,\at x=-1 is y=-7x-5