- #1
brizer
- 8
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1. The problem: The function f is continuous for -3[tex]\leq[/tex]c[tex]\leq[/tex]5 and differentiable for -3<x<5. If f(-3)=6 and f(5)=-2, which of the following could be false?
(a) there exists c, where -3[tex]\leq[/tex]c[tex]\leq[/tex]5, such that f(c)[tex]\geq[/tex]f(x) for all on the closed interval -3[tex]\leq[/tex]x[tex]\leq[/tex]5.
(b) There exists c, where -3[tex]\leq[/tex]c[tex]\leq[/tex]5, such that f'(c)=-1
(c) There exists c, where -3[tex]\leq[/tex]c[tex]\leq[/tex]5, such that f(c)=-1
(d) There exists c, where -3[tex]\leq[/tex]c[tex]\leq[/tex]5, such that f'(c)=0
(e) There exists c, where -3[tex]\leq[/tex]c[tex]\leq[/tex]5, such that f(c)=0
2. Homework Equations : IVT, mean value theorem
3. The Attempt at a Solution : If f(-3)=6 and f(5)=-2, then folloing the IVT, there is a c such that f(c)=-1 and f(c)=0. Following the mean value theorem, (-2-6)/(6--2)=-8/8=-1=f'(c). That leaves options (a) and (d). The only other theorem I can think of relevant to closed intervals of continuous, differentiable functions is Rolle's theorem which is not relevant. I feel like option (a) must be a theorem I can't remember, but I couldn't find it in my book, so I'm not sure.
(a) there exists c, where -3[tex]\leq[/tex]c[tex]\leq[/tex]5, such that f(c)[tex]\geq[/tex]f(x) for all on the closed interval -3[tex]\leq[/tex]x[tex]\leq[/tex]5.
(b) There exists c, where -3[tex]\leq[/tex]c[tex]\leq[/tex]5, such that f'(c)=-1
(c) There exists c, where -3[tex]\leq[/tex]c[tex]\leq[/tex]5, such that f(c)=-1
(d) There exists c, where -3[tex]\leq[/tex]c[tex]\leq[/tex]5, such that f'(c)=0
(e) There exists c, where -3[tex]\leq[/tex]c[tex]\leq[/tex]5, such that f(c)=0
2. Homework Equations : IVT, mean value theorem
3. The Attempt at a Solution : If f(-3)=6 and f(5)=-2, then folloing the IVT, there is a c such that f(c)=-1 and f(c)=0. Following the mean value theorem, (-2-6)/(6--2)=-8/8=-1=f'(c). That leaves options (a) and (d). The only other theorem I can think of relevant to closed intervals of continuous, differentiable functions is Rolle's theorem which is not relevant. I feel like option (a) must be a theorem I can't remember, but I couldn't find it in my book, so I'm not sure.