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

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

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

tangent of

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

y = 3 x + 2

Solution steps

Find the tangent point:

(- 1, - 1)

Find the slope of

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

Compute the slope of

f (x) = x^{3}

x = - 1 : m = 3

Find the line with slope m=

3

and passing through

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

y = 3 x + 2

\frac{d}{d x} (\frac{24}{( 1 + x ) ^{5}})

y^{'} = y^{2} - 2 y

\int (5 x + 3)^{3} d x

\int \frac{( x ^{2} + 1 )}{( x - 10 ) ( x - 9 ) ^{2}} d x

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

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