What is the tangent of f(x)=\sqrt[3]{x},\at x=64 ?

The tangent of f(x)=\sqrt[3]{x},\at x=64 is y= 1/48 x+8/3

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tangent of f(x)=\sqrt[3]{x},\at x=64

tangent of

f (x) = 3 x, at x = 64

y = \frac{1}{48} x + \frac{8}{3}

Solution steps

Find the tangent point:

(64, 4)

Find the slope of

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

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}{48}

Find the line with slope m=

\frac{1}{48}

and passing through

(64, 4) : y = \frac{1}{48} x + \frac{8}{3}

y = \frac{1}{48} x + \frac{8}{3}

\int 2 x^{2} + 5 d x

\frac{x d y}{d x} = - y + 2 x^{2} y, y (1) = 1

\frac{d}{d x} (9 x - 8 x)

y \to 2 lim (4 y^{2} - 2 y)

\int_{0}^{- 0.3} 300 x d x

What is the tangent of f(x)=\sqrt[3]{x},\at x=64 ?
The tangent of f(x)=\sqrt[3]{x},\at x=64 is y= 1/48 x+8/3