Find a parameterization for the curve y=8−4x3 that passes through the point (0,8,4) when t=−1 and is parallel to the xy-plane.
1 answer:
You might be interested in
<u>Answer:</u>
-4
<u>Step-by-step explanation:</u>
We are given the following expression which we are to solve:
Since this expression includes brackets as well has plus and minus signs, so we will follow the standard order of operations to solve this expression.
Solving the brackets first to get:
Now adding all the terms up to get:
-4
Answer:
Step-by-step explanation:
The angles of a triangle add to 180, add the angles and solve for x
9514 1404 393
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.
_____
<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.
