Consider the function X^3-2X-68. This function is zero when X is between 4 and 5. How do I know? If we ignore the 2X term in the function because it's going to be smaller than X^3 if X>1, X is approximately 4, since 4 cubed is close to 68 at 64. Another way to prove this is to draw the graph of the function roughly and see where it crosses the X-axis. Yet another way is to roughly plot f(X)=2X (a straight line) g(X)=X^3-68 (a curve) on the same graph and inspect where they intersect. At X=4, f(X)=8 while g(X)=-4, and at X=5, f(X)=10 while g(X)=57. In the first case the line is above the curve and in the second case it's below the curve, so the line and curve intersect somewhere between the two values of X.
My calculator allows me to input a single-variable function, but not solve it. If I input a range for the variable and an increment it tabulates the value of the function for all values in the range, using the increment to step through each value. If you don't have such a calculator, you just need to repeat calculations progressively noting the results as you go along. Here's what my calculator gave me for different values of X.
| 4.0 |
-12 |
| 4.1 |
-7.279 |
| 4.2 |
-2.312 |
| 4.3 |
2.907 |
| 4.4 |
8.384 |
| 4.5 |
14.125 |
| 4.6 |
20.136 |
| 4.7 |
26.423 |
| 4.8 |
32.992 |
| 4.9 |
39.849 |
As you can see the sign changes from - to + between the values X=4.2 and 4.3. To be sure that this first decimal place is accurate I took the function to the next level, and this time the values were X=4.24 and 4.25, so the answer is nearer to 4.2. You would need to calculate the function for X=4, 4.1, 4.2 and 4.3 to observe the sign change. You could then move on to trying 4.21, 4.22, 4.23, 4.24 and 4.25 to be sure you were accurate to one decimal place.
