What is the extreme f(x)=1+x-sqrt(3)x^2 ?

The extreme f(x)=1+x-sqrt(3)x^2 is Maximum(1/(2sqrt(3)), 1/(4sqrt(3))+1)

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extreme f(x)=1+x-sqrt(3)x^2

extreme points

f (x) = 1 + x - 3 x^{2}

Maximum (\frac{1}{2 3}, \frac{1}{4 3} + 1)

Solution steps

f^{'} (x) = - 23 x + 1

Find intervals:

Increasing

: - \infty < x < \frac{1}{2 3},

Decreasing

: \frac{1}{2 3} < x < \infty

Plug

x = \frac{1}{2 3}

into

1 + x - 3 x^{2} : \frac{1}{4 3} + 1

Maximum (\frac{1}{2 3}, \frac{1}{4 3} + 1)

f (x) = \frac{x ^{2}}{2 x + 4}

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

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

f (x) = \frac{x}{x ^{2} - x + 1}, 0 \leq x \leq 3

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

What is the extreme f(x)=1+x-sqrt(3)x^2 ?
The extreme f(x)=1+x-sqrt(3)x^2 is Maximum(1/(2sqrt(3)), 1/(4sqrt(3))+1)