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

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

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

tangent of

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

y = 3 x + 3

Solution steps

Find the slope of

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

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = 3

Find the line with slope m=

3

and passing through

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

y = 3 x + 3

3 (x + 1), 2 (x + 1), 0, 7

f (x) = \frac{2 x + 14}{x + 9}

(2, - 2), (4, 1)

f (x) = \frac{5 x - 2}{5 x}

\int (- x) d x

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