Polynomials

Hong Kong

Stage 4 - Stage 5

The function $f\left(x\right)=x^3-x^2-3x+15$`f`(`x`)=`x`3−`x`2−3`x`+15 has been graphed. Starting with the initial approximation $x_0=-2$`x`0=−2, we want to use two applications of Newton's method to find a better approximation to the single $x$`x`-intercept.

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a

Evaluate $f\left(x_0\right)$`f`(`x`0)

b

Evaluate $f'\left(x_0\right)$`f`′(`x`0)

c

Starting with an initial approximation of $x_0=-2$`x`0=−2, use one application of Newton's method to find a better approximation, $x_1$`x`1, correct to 2 decimal places if necessary.

d

Starting with $x_1=-2.69$`x`1=−2.69, use a second application of Newton's method to find a better approximation, $x_2$`x`2, correct to 2 decimal places if necessary.

e

Why would $x_0=0$`x`0=0 not be a good initial approximation for the $x$`x`-intercept?

It is on the wrong side of the $x$`x`-intercept.

A

Being near a turning point, the next approximation it would generate would be further from the $x$`x`-intercept rather than closer to it.

B

Easy

10min

The function $f\left(x\right)=3x+7\ln x$`f`(`x`)=3`x`+7`l``n``x` has a zero near $x_0=1$`x`0=1. We want to use one application of Newton's method to find a better approximation to this zero.

Easy

3min

The function $f\left(x\right)=2\tan x+\ln x$`f`(`x`)=2`t``a``n``x`+`l``n``x` has a zero near $x_0=2$`x`0=2. We would like to use one application of Newton's method to find a better approximation to this zero.

Easy

2min

Consider the equation $x^2-13=0$`x`2−13=0. Starting with the initial approximation $x_0=3.8$`x`0=3.8, we want to use one application of Newton's method to find a better approximation to the positive root of the equation.

Easy

4min

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