What method are you supposed to use.
Your cubic has one real root (a zero) at
x = appr. -1.116 plus 2 imaginary roots.
So I guess a real root lies between -1 and -2
check:
f(-1) = -1 + 2 + 1 - 5 = -3
f(-2) = 8 + 8 + 2 -5 = + 13
so the curve must have crossed the x-axis (the zero) somewhere between - 1 and -2
Where you supposed to try different f(x) 's ?
Determine between which consecutive integers one or more real zeros of f(x) = –x3 + 2x2 – x – 5 are located
3 answers
A third degree equation? I think I would graph it on your graphing calculator, see where it crosses the x axis.
Bob,
check my reply to your email from yesterday.
It applies to this question.
check my reply to your email from yesterday.
It applies to this question.