PRECAL MATH HW CHAPTER 6

Question

Approximate the real zero between -1 and 1 accurate to three places for the polynomial 𝑓 (𝑥) = 𝑥^4 − 3𝑥^3 + 1 using the Intermediate Value Theorem as described in these notes.

Answer 1

Since we don't know what your notes are, we cannot help you approximate the real zero according to the method in your notes. But, if I were to do it, here is what I would do:

Iteration Step #1:

f(-1) = 5 meaning the point (-1, 5) is on the graph of f(x)

f(1) = -1 meaning the point (1, -1) is on the graph of f(x)

Making a line between these points:

m = (5 - -1)/(-1 - 1) = 6/-2 = -3

Equation of the line: y - 5 = -3(x - -1) or y = -3x + 2

Using y = -3x + 2, estimate the zero:

0 = -3x + 2

-3x = -2

x = 2/3

So x = 0.667 this is the first approximation of the zero

Iteration Step #2:

Find f(2/3)

f(2/3) = 25/81 meaning the point (2/3, 25/81) is on the graph of f(x)

Next, using the points (2/3, 25/81) and (1, -1) find the equation of the line connecting them:

m = (25/81 - -1)/(2/3 - 1) = -106/27

Equation of the line: y - -1 = -106/27(x - 1) or y = -106/27x + 79/27

Using y = -106/27x + 79/27, estimate the zero:

0 = -106/27x + 79/27

-106/27x = -79/27

x = 79/106 = 0.7452830189

So x = 0.745 is the 2nd approximation of the zero

Iteration Step #3:

Find f(79/106)

f(79/106) = 0.0666259684

Next, using the points (0.7452830189, 0.0666259684) and (1, -1) find the equation of the line connecting them:

m = (0.0666259684 - -1)/(0.7452830189 - 1) = -4.187494543

Equation of the line: y - -1 = -4.187494543(x - 1) or y = -4.187494543x + 3.187494543

Using y = -4.187494543x + 3.187494543, estimate the zero:

0 = -4.187494543x + 3.187494543

-4.187494543x = -3.187494543

x = 0.7611937187

So x = 0.761 is the 3rd approximation of the zero

Iteration Step #4:

Find f(0.7611937187)

f(0.7611937187) = 0.0125795798

Next, using the points (0.7611937187, 0.0125795798) and (1, -1) find the equation of the line connecting them:

m = (0.0125795798 - -1)/(0.7611937187 - 1) = -4.240171466

Equation of the line: y - -1 = -4.240171466(x - 1) or y = -4.240171466x + 3.240171466

Using y = -4.240171466x + 3.240171466, estimate the zero:

0 = -4.240171466x + 3.240171466

x = 0.7641604807

So x = 0.764 is the 4th approximation of the zero

Iteration Step #5:

Find f(0.7641604807)

f(0.7641604807) = 0.0023129155

Next, using the points (0.7641604807, 0.0023129155) and (1, -1) find the equation of the line connecting them:

m = (0.0023129155 - -1)/(0.7641604807 - 1) = -4.249978621

Equation of the line: y - -1 = -4.249978621(x - 1) or y = -4.249978621x + 3.249978621

Using y = -4.249978621x + 3.249978621, estimate the zero:

0 = -4.249978621x + 3.249978621

x = 0.7647046987

So x = 0.765 is the 5th approximation of the zero.

Iteration Step #6:

Find f(0.7647046987)

f(0.7647046987) = 0.0004231685891

Next, using the points (0.7647046987, 0.0004231685891) and (1, -1) find the equation of the line connecting them:

m = (0.0004231685891 - -1)/(0.7647046987 - 1) = -4.251777078

Equation of the line: y - -1 = -4.251777078(x - 1) or y = -4.251777078x + 3.251777078

Using y = -4.251777078x + 3.251777078, estimate the zero:

0 = -4.251777078x + 3.251777078

x = 0.7648042261

So x = 0.765 is the 6th approximation of the zero. Notice that this is the same as the 5th approximation. So x = 0.765 is our 3 decimal estimate of the zero.

Answer 2

The Intermediate Value Theorem guarantees a zero between two points if there is a sign change with the function value between the two points. It does not provide the location of the zero.

f(1) = -1 and f(-1) = 5, so there is a sign change indicating there is a zero between -1 and 1 for the given function.

You can find the zero with a TI84 or you can graph the function on Desmos and find the zero at (0.765, 0), which is accurate to three decimal places as required in the problem.

Answer 3

x^4 - 3x^3 + 1 = 0

by descartes rule of signs there are maximum 2 real positive zeros

1 is between -1 and 1, the other is close to 3

there are no negative real zeros

there are 2 imaginary zeros

a 4th degree function has 4 zeros

the zero between -1 and 1 is close to x=1 slightly less than 1

rough guess might be about 0.8

as f(1) = -1

and f(-1) = 5

f(2) = -7

f(3) = 1

2nd zero might be about 3.0

actual zeros are closer to

0.76482 and 2.96149

accurate to 3 decimals, the zero between -1 and 1

is: 0.765