Find range of function y=1/2x+3 when domain =-2,0,2,4
1 answer:
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Answer:
55.5000143456...
Step-by-step explanation:
Use the formula for the area of a circle and solve for r:
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.
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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:
A. y = x + 10
Step-by-step explanation:
Given;
- The equation y = -7x - 5
- and another equation have a solution of (-2, 9)
If we equate the value of y in the first equation and the value of y in the second equation, then the solution of x should be -2 . Let's test this with equation A:
-7x - 5 = x + 10
-7x - x = 10 + 5
-x = 15
x = (15 × -2) ÷ 15 = -2
*<em>This proves that equation A has the same value of x as our first equation!</em>
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Let's test if equation A has a solution; y = 9:
y = (-2) + 10 = 9
*<em>This proves that equation A has the same value of y as our first equation!</em>
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So the second equation is,
A. y = x + 10
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