What is the tangent of f(x)=8x^{1/3}-5,\at x=8 ?

The tangent of f(x)=8x^{1/3}-5,\at x=8 is y= 2/3 x+17/3

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tangent of f(x)=8x^{1/3}-5,\at x=8

tangent of

f (x) = 8 x^{\frac{1}{3}} - 5, at x = 8

y = \frac{2}{3} x + \frac{17}{3}

Solution steps

Find the tangent point:

(8, 11)

Find the slope of

f (x) = 8 x^{\frac{1}{3}} - 5 : \frac{df ( x )}{d x} = \frac{8}{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{2}{3}

Find the line with slope m=

\frac{2}{3}

and passing through

(8, 11) : y = \frac{2}{3} x + \frac{17}{3}

y = \frac{2}{3} x + \frac{17}{3}

What is the tangent of f(x)=8x^{1/3}-5,\at x=8 ?
The tangent of f(x)=8x^{1/3}-5,\at x=8 is y= 2/3 x+17/3