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anonymous
 one year ago
Use the Intermediate Value Theorem to show that there is a root of the given equation in the specified interval: x^4+x3=0, (1,2)
anonymous
 one year ago
Use the Intermediate Value Theorem to show that there is a root of the given equation in the specified interval: x^4+x3=0, (1,2)

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ok the interval is (1,2) so they are looking for you to use the Int Value Thm.... Do you know what this theorem means?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Try and explain it to me and I will see if you nail it.... formulas are important but understanding is more so

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ok well my teacher gave me the theorem and said if f is continuous on an interval and N is any number between f(a) and f(b) then there exists C on (a,b) such that f(c)=N. Not sure I understand that though....

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ok let discuss then.... the easiest way is to think in terms of what happens when you use it to find a zero of a polynomial... like your problem for instance

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Since f is continuous there are no holes, jumps, gaps, kinks, etc.... so it can be represented by a smooth curve

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0now lets take the interval from (a,b) and stipulate that f(a) is positive and f(b) is negative (i.e. f(a)>0 and f(b)<0)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Since f is continuous the graph of the function must have crossed the xaxis somewhere inside the interval in order for the value of the function to change signs

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1442891184392:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Sorry its kind of messy, but this picture illustrates this point

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Now this relies specifically on the fact the end points are positive and negative... so we KNOW that the function must have crossed the xaxis AT LEAST once (though it could in principle cross way more than once)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Does this make sense?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yes thank you...I get all of that. What I don't understand now is how this applies to the problem

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh well that is why I chose the example I did.... in that definition you stated the same thing is going on only I just chose to pick the value of N to be zero

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So the goal of the question is for you to evaluate the given function with the two end points of the interval

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0You should find that one side is negative and one side is positive.... Since the function is continuous on that interval... the only way the function could change signs (from positive to negative) is if it crosses the xaxis.... i.e. has a zero

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0in this case c would correspond to the (one of but not all!!) zeros and hence f(c)=N=0

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0correspond to the zero* (singular)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0How am I supposed to get that zero? Plugging in 1 to the equation?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0This is a fourth order polynomial so it could have up to 4 real roots... but this mean value trick with this specified interval will only give us 1... we could repeat the process, but with only one interval we only have 1 shot to identify potentially 1 zero

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0No the equation (aka the function f(x)) isnt supposed to equal zero.... the interval you were given (1,2) is specified by its endpoints 1 and 2

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0evaluate f(1) and f(2) and show that one of them is positive and one of them is negative

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the only way this can happen AND f(x) be a continuous function is if f crosses the xaxis at some point (we dont know where) inside of this interval

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0this point that it crosses the xaxis will correspond to the c you had in your above definition and the "yvalue" (aka the value of the function) N will equal zero (we define it to be that since we are asked to look for a zero

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Does this make sense?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yes thank you so much for your patience and help!
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