What is the tangent of f(x)=x^3-2x+1,\at x=0 ?

The tangent of f(x)=x^3-2x+1,\at x=0 is y=-2x+1

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tangent of f(x)=x^3-2x+1,\at x=0

tangent of

f (x) = x^{3} - 2 x + 1, at x = 0

y = - 2 x + 1

Solution steps

Find the tangent point:

(0, 1)

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 = 0 : m = - 2

Find the line with slope m=

- 2

and passing through

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

y = - 2 x + 1

\frac{8}{( 1 - x ) ^{2}}

\int \frac{3 ( x + 2 )}{( x + 7 ) ^{4}} d x

\frac{d}{d x} (cos^{2} (π x))

\int_{0}^{1} - \frac{5 y}{e ^{2 y}} d y

- x^{3} + x, x = - 1, y = - \frac{1}{2}

What is the tangent of f(x)=x^3-2x+1,\at x=0 ?
The tangent of f(x)=x^3-2x+1,\at x=0 is y=-2x+1