What is the extreme f(x)=100-1/4 (x^2+33x) ?

The extreme f(x)=100-1/4 (x^2+33x) is Maximum(-33/2 , 2689/16)

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extreme f(x)=100-1/4 (x^2+33x)

extreme points

f (x) = 100 - \frac{1}{4} (x^{2} + 33 x)

Maximum (- \frac{33}{2}, \frac{2689}{16})

Solution steps

f^{'} (x) = \frac{1}{4} (- 2 x - 33)

Find intervals:

Increasing

: - \infty < x < - \frac{33}{2},

Decreasing

: - \frac{33}{2} < x < \infty

Plug

x = - \frac{33}{2}

into

100 - \frac{1}{4} (x^{2} + 33 x) : \frac{2689}{16}

Maximum (- \frac{33}{2}, \frac{2689}{16})

f (x, y) = - 2 x^{2} + 5 y^{2} - 9 x + 9 y + 6

f (x) = x^{3} + y^{2} - 3 x - 4 y + 7

x^{5} - 15 x^{3} + 3

h (x) = \frac{θ x ^{4} + 3 θ x ^{3} + 3 θ x ^{2} + θ x}{x ^{3} + 2 x ^{2} + x}

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

What is the extreme f(x)=100-1/4 (x^2+33x) ?
The extreme f(x)=100-1/4 (x^2+33x) is Maximum(-33/2 , 2689/16)