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

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

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

tangent of

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

y = 3 x - 3

Solution steps

Find the tangent point:

(1, 0)

Find the slope of

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

Compute the slope of

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

x = 1 : m = 3

Find the line with slope m=

3

and passing through

(1, 0) : y = 3 x - 3

y = 3 x - 3

\frac{d}{d x} (a^{2} - x^{2} + a arcsin (\frac{x}{a}))

\int x - \frac{1}{2} x d x

\frac{\partial}{\partial y} (tan (2 + 2 x^{2} y^{4} z^{2}))

\frac{d}{d x} (\frac{x ^{2}}{10})

\int \frac{5}{x + 1} d x

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