Intermediate Value Theorem

Question

Suppose a polynomial function f has the following properties: f(2)=7 and f(3)=-1. What does the Intermediate Value Theorem tell us about the zeros of f? 

I can't remember what this theorem is or how to find the zeros of f.

Answer 1

Kari,

The intermediate value theorem says that if the function is continuous (doesn't jump around), and at some x=domain value (here, 2) the function value (range) is positive (here, 7), but at some other domain value (here, 3) the function value has gone negative (here, -1), SOMEWHERE in between the function had to cross the value zero. It did that because it had to go from positive to negative, smoothly, so it had to have the value zero at at least one point between (it could have crossed the x axis more times, too, you just can't tell from the limited data you have).

 

You can't in fact know exactly where that crossing point is (from the problem as stated). It *could* be at 2.875 (2+7/8), if it's a linear function. But it doesn't have to be. If you DO know what degree the polynomial of the function is, then you may know that the maximum number of zeros of the function (=roots of the function) is the degree (highest power of the polynomial) .... assuming only non-negative integer values of the polynomial terms of the function!

 

Additional complications: roots (zeros) of the function can be multiple (repeated, identical) and also complex numbers.... if you see a parabola that doesn't touch down to the x-axis, it has imaginary roots related to the intersection of (its reflection through its vertex) with the x-axis.

Answer 2

Kari,

 

The function f must be continuous which a polynomial is. You are given two values of x = 2 & 3 whose corresponding values y = f(x) are one positive and the other negative.  Since this function is continuous in the x-interval [2,3], there has to be at least one value of x=x' in this interval that corresponds to y = f(x') = 0. It can't cross from y>0 to y <0 otherwise. duh! so this point x' is a zero of f(x). If the function wiggles in this [2, 3] interval near y=0 axis, it may have multiple zeros, but the theorem, not knowing the function in detail, only guarantees one such zero. 

 

So your answer is, the theorem guarantees at least one zero in this interval, and you determine them all by finding all x values for which f(x) = 0.  In other words, solve the polynomial.

Answer 3

It means that somewhere between the values x = 2, and x = 3, the graph of f(x) = 0, or crosses the x axis. since the signs of f(x) change from a positive number to a negative number.

 

In other words, the zero of the function f(x) is between 2 and 3.  It appears that it may cross closer to there x = 3.

 

I helps to draw the two points for a visual.

 

I hope this helps.