What is the extreme f(x)=-7(2x^3-125/32)^2+1 ?

The extreme f(x)=-7(2x^3-125/32)^2+1 is Saddle(0,-108351/1024),Maximum(5/4 ,1)

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extreme f(x)=-7(2x^3-125/32)^2+1

extreme points

f (x) = - 7 (2 x^{3} - \frac{125}{32})^{2} + 1

Saddle (0, - \frac{108351}{1024}), Maximum (\frac{5}{4}, 1)

Solution steps

f^{'} (x) = - 168 x^{5} + \frac{2625 x ^{2}}{8}

Find intervals:

Increasing

: - \infty < x < 0,

Increasing

: 0 < x < \frac{5}{4},

Decreasing

: \frac{5}{4} < x < \infty

Plug

x = 0

into

- 7 (2 x^{3} - \frac{125}{32})^{2} + 1 : - \frac{108351}{1024}

Saddle (0, - \frac{108351}{1024})

Plug

x = \frac{5}{4}

into

- 7 (2 x^{3} - \frac{125}{32})^{2} + 1 : 1

Maximum (\frac{5}{4}, 1)

Saddle (0, - \frac{108351}{1024}), Maximum (\frac{5}{4}, 1)

f (x) = \frac{( x y ) ^{2} + 8000 x + 8000 y}{x y}

f (x, y) = \frac{1}{( x + y )}

f (x) = \frac{1}{1 - x ^{2}}

f (x) = \frac{x ^{2}}{x + 1}, - 2 \leq x \leq 2

f (x) = \frac{- x - 8}{10 x ^{2} - 40 x + 27}

What is the extreme f(x)=-7(2x^3-125/32)^2+1 ?
The extreme f(x)=-7(2x^3-125/32)^2+1 is Saddle(0,-108351/1024),Maximum(5/4 ,1)