What is the tangent of f(x)=\sqrt[3]{1+x},\at x=0 ?

The tangent of f(x)=\sqrt[3]{1+x},\at x=0 is y= 1/3 x+1

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tangent of f(x)=\sqrt[3]{1+x},\at x=0

tangent of

y = \frac{1}{3} x + 1

Solution steps

Find the tangent point:

(0, 1)

Find the slope of

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 = \frac{1}{3}

Find the line with slope m=

\frac{1}{3}

and passing through

(0, 1) : y = \frac{1}{3} x + 1

y = \frac{1}{3} x + 1

What is the tangent of f(x)=\sqrt[3]{1+x},\at x=0 ?
The tangent of f(x)=\sqrt[3]{1+x},\at x=0 is y= 1/3 x+1