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Use the intermediate value theorem to show that the polynomial P(x) has a real zero in the interval [2,3]. Approximate this zero to two decimal places

P(x)=2x^4-4x^3-9

Mar 11, 2020

#1
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Use the intermediate value theorem to show that the polynomial P(x) has a real zero in the interval [2,3]. Approximate this zero to two decimal places

P(x)=2x^4-4x^3-9

I do not know offhand what intermediate value theorem is but I expect I can answer this question.

\(P(2)=2*2^3-4*2^3-9\\ P(2)=16-32-9\\ P(2)=-25\\~\\ P(3)=2*3^4-4*3^3-9\\ P(3)=162-108-9\\ p(3)>0\)

Since P(2) is negative and P(3) is positive there must be a root between x=2 and x=3

Try P(2.5)

\(P(x)=2x^4-4x^3-9\\ P(2.5)=2*2.5^4-4*2.5^3-9\\ P(2.5)=6.625\\ P(2.5)>0\\ \)

So it must be between  2 and 2.5

The midpoint of 2 and 2.5 is 2.25

\(P(x)=2x^4-4x^3-9\\ P(2.25)=2*2.25^4-4*2.25^3-9\\ P(2.25)=-3.3 \\P(2.25)<0\)

So the root is between  2.25 and 2.50    The midpoint of these is  2.375

\(P(x)=2x^4-4x^3-9\\ P(2.375)=2*2.375^4-4*2.375^3-9\\ P(2.375)=1.047 \\P(2.375)>0\)

So the root is between 2.25 and 2.375. The average of these is  2.3125

\(P(x)=2x^4-4x^3-9\\ P(2.3125)=2*2.3125^4-4*2.3125^3-9\\ P(2.3125)=-1.27 \\P(2.3125)<0\)

So the root is between  2.3125 and 2.375  The average of these is  2.34375 which is approx 2.344

\(P(x)=2x^4-4x^3-9\\ P(2.344)=2*2.344^4-4*2.344^3-9\\ P(2.344)=-0.139 \\P(2.344)<0\)

So the root is between  2.344 and  2.375. The average of these is   2.3595  which is approx 2.360

\(P(x)=2x^4-4x^3-9\\ P(2.360)=2*2.360^4-4*2.360^3-9\\ P(2.360)=0.4638 \\P(2.360)>0\)

So the root is between  2.360 and  2.344. The average of these is   2.352

\(P(x)=2x^4-4x^3-9\\ P(2.352)=2*2.352^4-4*2.352^3-9\\ P(2.352)=0.159 \\P(2.352)>0\)

So the root is between  2.352 and  2.344. The average of these is   2.348

\(P(x)=2x^4-4x^3-9\\ P(2.348)=2*2.348^4-4*2.348^3-9\\ P(2.348)=0.00955 \\P(2.348)>0 \)

So the root is between  2.348 and  2.344. The average of these is   2.346

\(P(x)=2x^4-4x^3-9\\ P(2.346)=2*2.346^4-4*2.346^3-9\\ P(2.346)= -0.65 \\P(2.346)<0\)

So the root is between  2.346 and  2.348.

So to 2 decimal places the root is approx 2.35

What an effort. Maybe you were meant to use a quicker method.

Coding:

P(2)=2*2^3-4*2^3-9\\
P(2)=16-32-9\\
P(2)=-25\\~\\
P(3)=2*3^4-4*3^3-9\\
P(3)=162-108-9\\
p(3)>0

P(x)=2x^4-4x^3-9\\
P(2.5)=2*2.5^4-4*2.5^3-9\\
P(2.5)=6.625\\
P(2.5)>0\\

P(x)=2x^4-4x^3-9\\
P(2.25)=2*2.25^4-4*2.25^3-9\\
P(2.25)=-3.3
\\P(2.25)<0

P(x)=2x^4-4x^3-9\\
P(2.375)=2*2.375^4-4*2.375^3-9\\
P(2.375)=1.047
\\P(2.375)>0

P(x)=2x^4-4x^3-9\\
P(2.3125)=2*2.3125^4-4*2.3125^3-9\\
P(2.3125)=-1.27
\\P(2.3125)<0

P(x)=2x^4-4x^3-9\\
P(2.344)=2*2.344^4-4*2.344^3-9\\
P(2.344)=-0.139
\\P(2.344)<0

P(x)=2x^4-4x^3-9\\
P(2.360)=2*2.360^4-4*2.360^3-9\\
P(2.360)=0.4638
\\P(2.360)>0

P(x)=2x^4-4x^3-9\\
P(2.352)=2*2.352^4-4*2.352^3-9\\
P(2.352)=0.159
\\P(2.352)>0

P(x)=2x^4-4x^3-9\\
P(2.348)=2*2.348^4-4*2.348^3-9\\
P(2.348)=0.00955
\\P(2.348)>0

Mar 11, 2020