What is the extreme \sqrt[4]{x^4-81} ?

The extreme \sqrt[4]{x^4-81} is Minimum(-3,0),Minimum(3,0)

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extreme \sqrt[4]{x^4-81}

extreme points

Minimum (- 3, 0), Minimum (3, 0)

Solution steps

f^{'} (x) = \frac{x ^{3}}{( x ^{4} - 81 ) ^{\frac{3}{4}}}

Find intervals:

Decreasing

: - \infty < x < - 3,

Increasing

: 3 < x < \infty

Plug

x = - 3

into

Minimum (- 3, 0)

Plug

x = 3

into

Minimum (3, 0)

Minimum (- 3, 0), Minimum (3, 0)

f (x, y) = x^{2} + x y + 2 x + 4 y + 1

f (x) = x^{3} + 3 x^{2} + 2 x

f (x) = 2 x^{3} - 21 x^{2} + 60 x, 1 \leq x \leq 6

y^{3} + 6 x^{2} y - 6 x^{2} - 6 y^{2} + 3

T (x, y) = 100 + x (y + 1)^{2} - 6 x^{2} - 8 (y + 1)^{2}

What is the extreme \sqrt[4]{x^4-81} ?
The extreme \sqrt[4]{x^4-81} is Minimum(-3,0),Minimum(3,0)