What is the tangent of f(x)=x^3+2x,\at x=-1 ?

The tangent of f(x)=x^3+2x,\at x=-1 is y=5x+2

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tangent of f(x)=x^3+2x,\at x=-1

tangent of

f (x) = x^{3} + 2 x, at x = - 1

y = 5 x + 2

Solution steps

Find the tangent point:

(- 1, - 3)

Find the slope of

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

Compute the slope of

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

x = - 1 : m = 5

Find the line with slope m=

5

and passing through

(- 1, - 3) : y = 5 x + 2

y = 5 x + 2

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

f (x) = x ∣ x ∣

f (x) = 5 x - 4

\int_{\frac{1}{3}}^{5} 12 x ln (3 x) d x

\int tan^{2} (x co) s^{3} x d x

What is the tangent of f(x)=x^3+2x,\at x=-1 ?
The tangent of f(x)=x^3+2x,\at x=-1 is y=5x+2