What is the extreme (5-9t)^{9/2} ?

The extreme (5-9t)^{9/2} is Minimum(5/9 ,0)

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extreme (5-9t)^{9/2}

extreme points

(5 - 9 t)^{\frac{9}{2}}

Minimum (\frac{5}{9}, 0)

Solution steps

Find the critical points:

t = \frac{5}{9}

Domain of

(5 - 9 t)^{\frac{9}{2}} : t \leq \frac{5}{9}

Find intervals:

Decreasing

: - \infty < t < \frac{5}{9}

Plug the extreme point

x = \frac{5}{9}

into

(5 - 9 t)^{\frac{9}{2}} \Rightarrow y = 0

Minimum (\frac{5}{9}, 0)

f (x) = 5 x^{2} ln (4 x)

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

x^{3} - 48 x y + 64 y^{3}

f (x, y) = 2 x^{3} + 3 x^{3} y^{2} - 4 x^{2} y^{2} - 2 y

f (x) = e^{x} - 7 e^{- x} - 8 x