Using the Intermediate Value Theorem, approximate the real zero of f(x)= x^{3}+x^{2}1.
x 
f(x) 
2 

1 

0 

1 

2 

Now, using the IVT again, divide the interval you found above into tenths in order to pinpoint the zero more closely.This means that the zero of this function (approximated to the closest whole numbers) occurs between the values x= and x= .
The zero of this function (approximated now to the closest tenth) between the values of x= and x= .
Plugging in the xvalues given in the table for x, yields the following y values.
x 
f(x) 
2 
5 
1 
1 
0 
1 
1 
1 
2 
11 
We know then, that the zero of f(x) (approximated to the nearest whole numbers) occurs between x=0 and x=1 (since f(0) is negative and f(1) is positive)
To approximate the zero of this function to the closest tenths, divide the interval [0, 1] into tenths.
x 
f(x) 
0.0 
5 
0.1 
.989 
0.2 
.952 
0.3 
.883 
0.4 
.776 
0.5 
.625 
0.6 
.424 
0.7 
.167 
0.8 
.152 
0.9 
.539 
1.0 
1 
After evaluating the function at each point, find that f(0.7) is negative and f(0.8) is positive, meaning that the function must have a zero between x=0.7 and x=0.8.