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Continuity (intermediate value theorem)

  • Thread starter mtayab1994
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Homework Statement



Let n be a natural number.
Prove that the equation: [tex]x^{2}(cos(x))^{n}+xsin(x)+1=0[/tex] has an infinite amount of solutions.




The Attempt at a Solution



I named that equation f(x)=0 and I said that f(a)<f(0)<f(b) and that f(a) x f(b) < 0.

Should I choose n=1 and chose a=0 and b=pi ? I know that the equation is a period equation so it will have an infinite amount of solutions, but i'm not sure how to show it because I've only been dealing with equations that want you to show there is at least one solution in [a,b]. Any help would be really appreciated. Thank you.
 

Answers and Replies

  • #2
jbunniii
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Hint: consider the values of [itex]x[/itex] for which [itex]\cos(x) = 0[/itex]. What does [itex]\sin(x)[/itex] equal for these values of [itex]x[/itex]?
 
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Hint: consider the values of [itex]x[/itex] for which [itex]\cos(x) = 0[/itex]. What does [itex]\sin(x)[/itex] equal for these values of [itex]x[/itex]?
When cos(x)=0 then sin(x)=1 for values in the form of x= pi/2+2kpi

So then we could have : f(3pi/2) < f(0) < f(pi/2) ??
 
  • #4
jbunniii
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When cos(x)=0 then sin(x)=1 for values in the form of x= pi/2+2kpi
OK, and what about [itex]x = 3\pi/2 + 2k\pi[/itex]?

So then we could have : f(3pi/2) < f(0) < f(pi/2) ??
Are you asking me, or telling me? What does f(x) reduce to when [itex]x = \pi/2 + 2k\pi[/itex]? How about when [itex]x = 3\pi/2 + 2k\pi[/itex]?
 
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OK, and what about [itex]x = 3\pi/2 + 2k\pi[/itex]?


Are you asking me, or telling me? What does f(x) reduce to when [itex]x = \pi/2 + 2k\pi[/itex]? How about when [itex]x = 3\pi/2 + 2k\pi[/itex]?
Don't mind me I've had a really long day i wanted to say x=pi/2+kpi.
 
  • #6
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Ok this is what i got: f(pi/2)=pi/2+1 and f(3pi/2)=1-3pi/2 so we can say that:

f(3pi/2) < 0 < f(pi/2) am i correct?
 
  • #7
jbunniii
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Ok this is what i got: f(pi/2)=pi/2+1 and f(3pi/2)=1-3pi/2 so we can say that:

f(3pi/2) < 0 < f(pi/2) am i correct?
Yes, that's right. Now you can apply the intermediate value theorem.

You can argue similarly for [itex]f(3\pi/2 + 2k\pi)[/itex] and [itex]f(\pi/2 + 2k\pi)[/itex], for any positive integer [itex]k[/itex].
 
  • #8
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Yes, that's right. Now you can apply the intermediate value theorem.

You can argue similarly for [itex]f(3\pi/2 + 2k\pi)[/itex] and [itex]f(\pi/2 + 2k\pi)[/itex], for any positive integer [itex]k[/itex].
Alright I've got it thanks for your help.
 

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