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3-3(0)^2<= g(0)<= 3cos(0)

Solution

3 - 3 (0)^{2} \leq g (0) \leq 3 cos (0)

No Solution

Solution steps

3 - 3 (0)^{2} \leq g (0) \leq 3 cos (0)

a \leq u \leq b

then

a \leq u and u \leq b

3 - 3 (0)^{2} \leq g (0) and g (0) \leq 3 cos (0)

3 - 3 \cdot 0^{2} \leq g \cdot 0 :

No Solution

3 - 3 \cdot 0^{2} \leq g \cdot 0

Switch sides

g \cdot 0 \geq 3 - 3 \cdot 0^{2}

Apply rule

0 \cdot a = 0

0 \geq 3 - 3 \cdot 0^{2}

Apply rule

0^{a} = 0

0^{2} = 0

0 \geq 3 - 3 \cdot 0

Apply rule

0 \cdot a = 0

0 \geq 3 - 0

Subtract the numbers:

3 - 0 = 3

0 \geq 3

Therefore, the solution is

No Solution

g \cdot 0 \leq 3 cos (0) :

True for all

g

g \cdot 0 \leq 3 cos (0)

Simplify

g \cdot 0 : 0

g \cdot 0

Apply rule

0 \cdot a = 0

= 0

Simplify

3 cos (0) : 3

3 cos (0)

Simplify

cos (0) : 1

cos (0)

Use the following trivial identity:

cos (0) = 1

cos (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= 1

= 3 \cdot 1

Multiply the numbers:

3 \cdot 1 = 3

= 3

0 \leq 3

Therefore, the solution is

True for all g

Combine the intervals

No Solution and True for all g

Merge Overlapping Intervals

No Solution and True for all g

The intersection of two intervals is the set of numbers which are in both intervals

No Solution

and

True for all

g

No Solution

No Solution