Is any real number exactly 1 less than its cube? Answer this question by applying the intermediate value property (you need to determin a suitable finction and a suitable interval)
Is any real number exactly 1 less than its cube? Answer this question by applying the intermediate value property (you need to determin a suitable finction and a suitable interval)
I really dont understand, can someone help?
Thanks
You mean,
x=x^3-1
Rewrite as,
x^3 -x -1=0
When x=10, the polynomial is positive.
When x=-10, the polynomial is negative.
Thus, there exists at least one zero on the interval (-10,10).