The statement that the value of 3 is an upper bound for the zeros of the function shown is;
B: False
We are given the function;
f(x) = -3x³ + 20x² – 36x + 16.
In this function, we see that there are three sign changes. Thus by Descarte's rule of signs, there exist 3 or 1 positive zeros
Let's now check f(-x) ;
f(-x) = 3x³ + 20x² + 36x + 16.
There is no sign change in this case and by Descart’s rule of signs, there is no
negative root.
The leading coefficient in the polynomial is -3, and so we must divide all terms of the main polynomial by 3. This gives;
x³ - 6.67x² + 12x - 5.33
The coefficients are 1; -6.67; -12; -5.33
Drop the leading coefficient and remove any minus signs to get: 6.67; 12; 5.33
Bound 1: the largest value is 12. We add 1 to get 13
Bound 2: adding all values gives: 6.67 + 12 + 5.33 = 24
We see that the sum of all values is greater than 1.
The smallest “bounds” value is 13.
Now, all the roots are within ±13.
Thus, 3 is not an upper bound.
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