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Antwort 1.3:1
Aus Online Mathematik Brückenkurs 2
| a)
| Critical point: | x=0
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| Inflexion point: | None
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| Local min: | x=0
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| Local max: | None
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| Global min: | x=0
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| Global max: | None
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| Strictly incr.: | x 0
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| Strictly decr.: | x 0
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| b)
| Critical point: | x=−1, x=1
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| Inflexion point: | None
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| Local min: | x=−3, x=1
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| Local max: | x=−1, x=2
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| Global min: | x=−3
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| Global max: | x=−1
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| Strictly incr.: | [−3 −1], [12]
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| Strictly decr.: | [−11]
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| c)
| Critical point:
| \displaystyle x=-2, \displaystyle x=-1, \displaystyle x=\tfrac{1}{2}
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| Inflexion point: | \displaystyle x=-1
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| Local min: | \displaystyle x=-2, \displaystyle x=2
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| Local max: | \displaystyle x=-3, \displaystyle x=\tfrac{1}{2}
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| Global min: | \displaystyle x=-2
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| Global max: | \displaystyle x=-3
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| Strictly incr.: | \displaystyle [-2,\tfrac{1}{2}]
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| Strictly decr.: | \displaystyle [-3,-2], \displaystyle [\tfrac{1}{2},2]
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| d)
| Critical point:
| \displaystyle x=-\tfrac{5}{2}, \displaystyle x=\tfrac{1}{2}
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| Inflexion point: | None
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| Local min:
| \displaystyle x=-\tfrac{5}{2}, \displaystyle x=-\tfrac{1}{2}, \displaystyle x=2
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| Local max:
| \displaystyle x=-3, \displaystyle x=-1, \displaystyle x=\tfrac{1}{2}
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| Global min: | \displaystyle x=-\tfrac{5}{2}
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| Global max: | \displaystyle x=-1
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| Strictly incr.:
| \displaystyle [-\tfrac{5}{2},-1], \displaystyle [-\tfrac{1}{2},\tfrac{1}{2}]
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| Strictly decr.:
| \displaystyle [-3,-\tfrac{5}{2}], \displaystyle [-1,-\tfrac{1}{2}], \displaystyle [\tfrac{1}{2},2]
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