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

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

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

tangent of

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

y = x + 3

Solution steps

Find the tangent point:

(- 1, 2)

Find the slope of

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

Compute the slope of

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

x = - 1 : m = 1

Find the line with slope m=

1

and passing through

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

y = x + 3

(e^{x} - e^{- x}) \frac{d y}{d x} = y^{2}

\frac{\partial}{\partial x} (- 10 x + 7 y^{5})

\frac{d}{d x} (\frac{1}{4 x ^{3}} - \frac{6}{5 x ^{4}})

\frac{d y}{d x} = (\frac{2 y + 5}{8 x + 7})^{2}

y^{^{'}} - 3 y = 3 e^{4 t}

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