[1] ROOTS OF THE FUNCTION:
f(x) = (x+1/4)^4 (x^2+1)^3
0 = (x+1/4)^4
x = -1/4
f(x) = (x+1/4)^4 (x^2+1)^3
0= (x^2+1)^3
x^2 = -1
x = √(-1)
x = i
[2] GRAPH:
Does the graph touch the axis?
f(0) = (0+1/4)^4 (0^2+1)^3
Graph Touches the graph at: f(0)
f(0) = (1/4)^1/4 +1 ~ 1.707
2006-07-10 17:18:51
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answer #1
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answered by Ernest Maxwell 2
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Set each factor = to 0.
x+1/4 =0 > x= - 1/4 Multiplicity is 4 (power on the factor)
X^2 + 1 = 0 > x^2 = -1 > x = +/- i (imaginary solution, i and -i each have multiplicity 3)
The graph just touches the x-axis at 1/4, since that factor has an even power.
2006-07-10 16:20:33
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answer #2
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answered by jenh42002 7
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(x + 1/4) = 0, x = -1/4, but since it was power 4, it has multiplicity 4. The graph just touches the x-axis because for even numbered multiplicity, the graph only touches the "zero"
(x^2 + 1) = 0 x = sqrt(-1), with power 3, it has multiplicity 3, but it is imaginary, so it doesn't cross the real axes.
(Normally, odd numbered multiplicities involve the graph crossing the x-axis, thus changing sign)
2006-07-10 16:27:01
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answer #3
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answered by Anonymous
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x + 1/4 = 0
x = -1/4
This has a multiplicity of 4 (because that is the exponent on (x+1/4))
It touches the x-axis because the multiplicity is even.
(Note: It crosses the x-axis when the multiplicity is odd.)
x^2 + 1 = 0
x^2 = -1
This has imaginary solutions.
(I assume you are only looking for real solutions.)
2006-07-10 16:14:54
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answer #4
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answered by MsMath 7
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it really is a quadratic equation: x^2 - 7x = -13 [Make the equation equivalent 0] x^2 - 7x + 13 = 0 [Now, use the quadratic formulation] ax^2 + bx + c =0 x = (-b +/- (b^2 - 4ac)^one million/2)/(2a) x= (Imaginary Numbers) There are not any authentic techniques to the equation; unsolvable to the quantity of authentic techniques. The quadratic formulation is genuine on mine; the variable ^one million/2 is an similar as sq. root. I in basic terms did not favor to discover the sq. root image.
2016-12-10 07:43:30
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answer #5
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answered by ? 4
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I know, but I'm not going to tell you. (Yeah, right!) Good luck!
2006-07-10 16:16:37
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answer #6
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answered by NannyMcPhee 5
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do ur own homework.
2006-07-10 16:14:17
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answer #7
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answered by Rabi 3
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