Intermediate Value Theorem

Problem: Use the Intermediate Value Theorem to show that the following function has a zero in the given interval.  Approximate the zero to two decimal places.

f(x)=9x^3+9x^2-9x+6; [-2,-1]

Solution:

To determine if there is a zero in the interval use the Intermediate Value theorem.  To use the Intermediate Value Theorem, the function must be continuous on the interval [-2,-1].  The function f(x)=9x^3+9x^2-9x+6 is a polynomial function and polynomial functions are defined and continuous for all real numbers.

Evaluate the function at the endpoints and if there is a sign change.  If there is a sign change, the Intermediate Value Theorem states there must be a zero on the interval.  To evaluate the function at the endpoints, calculate f(-2)  and f(-1).

f(-2)=9(-2)^3+9(-2)^2-9(-2)+6
f(-2)=9(-8)+9(4)+18+6
f(-2)=-72+36+24
f(-2)=-72+60
f(-2)=-12

f(-1)=9(-1)^3+9(-1)^2-9(-1)+6
f(-1)=9(-1)+9(1)+9+6
f(-1)=-9+9+9+6
f(-1)=0+15
f(-1)=15

Since one endpoint gives a negative value and one endpoint gives a positive value, there must be a zero in the interval.

We can get a better approximation of the zero by trying to figure out the next decimal point. Write out all of the values to one decimal point between -2 and -1.

Fill the table.  There are functions in your calculator that make this easier.

Use the Intermediate Value Theorem again.  Look for a sign change.  Looking down the table, there is a sign change between -1.8 and -1.7.  With this information we now know the zero is between these two values.

Repeat this process again with two decimal places between -1.8 and -1.7.

Use the Intermediate Value Theorem.  Look for a sign change.  Looking down the table, there is a sign change between -1.78 and -1.77.  With this information we now know the zero is between these two values and the zero to two decimal places is -1.77 since all the numbers between -1.78 and -1.77 start with -1.77.