To approximate the real number solution, we will first evaluate the function at various points to find a root using the Intermediate Value Theorem.
Let’s evaluate the function at some integer values:
f(1) = 1³ – 2(1)² + 5(1) – 6 = 1 – 2 + 5 – 6 = -2
f(2) = 2³ – 2(2)² + 5(2) – 6 = 8 – 8 + 10 – 6 = 4
Since f(1) is negative and f(2) is positive, there is at least one root between 1 and 2.
We can narrow it down further:
f(1.5) = (1.5)³ – 2(1.5)² + 5(1.5) – 6 = 3.375 – 4.5 + 7.5 – 6 = 0.375
Here, f(1) is still negative and f(1.5) is positive, so the root is between 1 and 1.5.
Next, we try:
f(1.25) = (1.25)³ – 2(1.25)² + 5(1.25) – 6 = 1.953125 – 3.125 + 6.25 – 6 = -1.921875
Now, we see f(1.25) is negative and f(1.5) is positive, so the root is between 1.25 and 1.5.
Continuing this process will help us hone in on a more precise value, but as a rough estimate, we can conclude that the approximate real number solution s is around 1.3.