90.
use intermediate value theorem and the table to find intervals 1 unit in the length in which the polynomial is quarantined to have a zero
In other words, the Intermediate Value Theorem tells us that when a polynomial function changes from a negative value to a positive value, the function must cross the x-axis.
View attachment 353
Let f be a polynomial function. The Intermediate Value Theorem states that if f(a) and f(b) have opposite signs, then there exists at least one value c between a and b for which f(c)=0.
As a start, evaluate f(x) at the integer values x=0,1,2,3,4,......,
f(x)=0.11x^3-2.07x^2+9.81x-6.88
first graph
table ( choosing x observing the graph)
x| f(x)
0| -6.88
1| 0.97.......................f(1)=0.11(1)^3-2.07(1)^2+9.81(1)-6.88=0.97
6| 1.22.......................f(6)=0.11(6)^3-2.07(6)^2+9.81(6)-6.88=1.22
7| -1.91.......................f(7)=0.11(7)^3-2.07(7)^2+9.81(7)-6.88=-1.91
11| -3.03.......................f(11)=0.11(11)^3-2.07(11)^2+9.81(11)-6.88=-3.03
12| 2.84.......................f(12)=0.11(12)^3-2.07(12)^2+9.81(12)-6.88=2.84
[0,1]->This interval does work because as x increases, f(x) changes signs. There is a zero.
[6,7]->This interval does work because as x increases, f(x) changes signs. There is a zero.
[11,12]->This interval does work because as x increases, f(x) changes signs. There is a zero.
We have shown that there are at least two real zeros between x=0 and x=1, x=6 and x=7, x=11 and x=12
b.
adjust the table to approximate the zeros of the function to the nearest thousandth
let's check x=0.839 , x=6.345, and x=11.612
x|f(x)
0.839| -0.042.......................f(0.839)=0.11(0.839)^3-2.07(0.839)^2+9.81(0.839)-6.88=-0.301->round to 0
6.345| 0.269.......................f(6.345)=0.11(6.345)^3-2.07(6.345)^2+9.81(6.345)-6.88=0.127->round to 0
11.612| 0.18.......................f(11.612)=0.149->round to 0
approximate zeros are: x=0.839 , x=6.345, and x=11.612
find actual zeros:
easiest way is to
factor
f(x)=0.11x^3-2.07x^2+9.81x-6.88.......
factor completely
f(x) = 0.11 (x -
11.5878) (x -
6.38506) (x -
0.845338)
zeros are at:
x≈0.845
x≈6.385
x≈11.588