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Answer:
m ≥ -√5
Step-by-step explanation:
If g(x) = f(|x|) for x∈i, then g(x) = f(x) for x∈ℝ: x ≥ 0.
f'(x) is 5th-degree, so f(x) is 6th-degree, meaning it is generally U-shaped. Since we're only concerned with x ≥ 0, we want to make sure f'(x) has no real zeros of odd multiplicity such that x > 0. The given factors of f'(x) make it have real zeros at x = -3 and x = -1.
For the last factor, (x² +2mx +5) to have no positive real zeros of odd multiplicity, we must have m ≥ 0 or the discriminant ≤ 0. The discriminant is ...
d = b² -4ac = (2m)² -4(1)(5) = 4m² -20 . . . . . discriminant of the last factor
d ≤ 0 . . . . . . . . . . the condition for no real zeros
4m² -20 ≤ 0
m² -5 ≤ 0 . . . . . . divide by 4
m² ≤ 5 . . . . . . . . .add 5
|m| ≤ √5 . . . . . . . take the square root
This tells us there will be a positive real zero of multiplicity 2 in f'(x) when m = -√5, and there will be no positive real zeros for -√5 < m < 0
There will be no odd-multiplicity positive real zeros in the derivative function f'(x) as long as m ≥ -√5. This means the slope of f(x) is non-negative for x ≥ 0, hence f(|x|) has its only minimum at x=0.
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<em>Additional comment</em>
The multiplicity of the zeros of f'(x) is important because the derivative will only change sign where the multiplicity is odd. When the discriminant of (x²+2mx+5) is zero, the associated positive real zero will have multiplicity 2, hence f'(x) will not change sign there.
Answer:
There are <u>20,000 milligrams</u> chlorine in the pool.
Step-by-step explanation:
Given:
The chlorine per liter of water in the pool is 2.5 milligrams.
The volume of water in the swimming pool is 8000 gallons.
Now, to find the quantity of chlorine in the pool.
Let the quantity of chlorine be
The gallons of water in the swimming pool = 8000.
The milligrams of chlorine per liter of water in the pool = 2.5.
According to question, the amount of chlorine in the swimming pool is proportional to the volume of water.
So, 2.5 mg of chlorine is equivalent to 1 liter.
Thus, is equivalent to
Now, to get the quantity of chlorine by using cross multiplication method:
<em>By cross multiplying we get:</em>
Therefore, there are 20,000 milligram chlorine in the pool.