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

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

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

tangent of

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

y = 5 x + 3

Solution steps

Find the tangent point:

(- 1, - 2)

Find the slope of

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

Compute the slope of

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

x = - 1 : m = 5

Find the line with slope m=

5

and passing through

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

y = 5 x + 3

\int x^{2} cos (x^{3} + 1) d x

\int (- \frac{4000}{x ^{2}}) d x

\frac{d}{d x} (3 e^{3 x} cos (2 x) - 2 e^{3 x} sin (2 x))

x^{2} d x + y (x - 1) d y = 0

4 + x^{6}

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