- anonymous

Use the Intermediate Value Theorem to show that there is a root of the given equation in the specified interval: x^4+x-3=0, (1,2)

- jamiebookeater

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- anonymous

Ok 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

Try and explain it to me and I will see if you nail it.... formulas are important but understanding is more so

- anonymous

Ok 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....

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- anonymous

Ok 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

Since f is continuous there are no holes, jumps, gaps, kinks, etc.... so it can be represented by a smooth curve

- anonymous

now 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

Since f is continuous the graph of the function must have crossed the x-axis somewhere inside the interval in order for the value of the function to change signs

- anonymous

|dw:1442891184392:dw|

- anonymous

Sorry its kind of messy, but this picture illustrates this point

- anonymous

Now this relies specifically on the fact the end points are positive and negative... so we KNOW that the function must have crossed the x-axis AT LEAST once (though it could in principle cross way more than once)

- anonymous

Does this make sense?

- anonymous

Yes thank you...I get all of that. What I don't understand now is how this applies to the problem

- anonymous

Oh 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

So the goal of the question is for you to evaluate the given function with the two end points of the interval

- anonymous

You 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 x-axis.... i.e. has a zero

- anonymous

in this case c would correspond to the (one of but not all!!) zeros and hence f(c)=N=0

- anonymous

correspond to the zero* (singular)

- anonymous

How am I supposed to get that zero? Plugging in 1 to the equation?

- anonymous

This 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

No 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

evaluate f(1) and f(2) and show that one of them is positive and one of them is negative

- anonymous

the only way this can happen AND f(x) be a continuous function is if f crosses the x-axis at some point (we dont know where) inside of this interval

- anonymous

this point that it crosses the x-axis will correspond to the c you had in your above definition and the "y-value" (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

Does this make sense?

- anonymous

Yes thank you so much for your patience and help!

- anonymous

No sweat dude :D

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