The value 3 is an upper bound for the zeros of the function shown below.

f(x)=-3x³+20x²-36x+16

a. true
b. false

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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