What is the tangent of-9x^{1/3}+5,\at x=27 ?

The tangent of-9x^{1/3}+5,\at x=27 is y=-1/3 x-13

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tangent of-9x^{1/3}+5,\at x=27

tangent of

- 9 x^{\frac{1}{3}} + 5, at x = 27

y = - \frac{1}{3} x - 13

Solution steps

Find the tangent point:

(27, - 22)

Find the slope of

y = - 9 x^{\frac{1}{3}} + 5 : \frac{d y}{d x} = - \frac{3}{x ^{\frac{2}{3}}}

Compute the slope of

y = - 9 x^{\frac{1}{3}} + 5

x = 27 : m = - \frac{1}{3}

Find the line with slope m=

- \frac{1}{3}

and passing through

(27, - 22) : y = - \frac{1}{3} x - 13

y = - \frac{1}{3} x - 13

\frac{d}{d x} (x^{2} - 2 x - \frac{2}{x ^{4}})

\int + (x^{2} - 1) d x

\int π^{2} d x

x \to 3 lim (x^{2} + 3)

What is the tangent of-9x^{1/3}+5,\at x=27 ?
The tangent of-9x^{1/3}+5,\at x=27 is y=-1/3 x-13