Polynomials

Hong Kong

Stage 4 - Stage 5

Lesson

When we want to find the roots of the equation $P(x)=0$`P`(`x`)=0, or the $x$`x`-intercepts (the zeros) of the function $y=P(x)$`y`=`P`(`x`), we have a number of strategies up our sleeve.

- If we have a nice quadratic or cubic function, we'll likely try to factorise it, or perhaps use the quadratic formula.
- If we have some technology at our disposal, then we can graph the function to find the zeros or use an algorithm (a certain computational process) to find the roots.

But how does this technology actually go about finding these roots? There's not a little smart dude sitting in there doing the algebra for you. Instead, your calculator or online software is programmed with various algorithms that calculate, at very high speeds, a process that finds the roots for you.

These processes are called Numerical Approximations. The calculator doesn't always give you an exact solution, but an extremely close approximation.

We're going to learn one of these processes today.

The best way to demonstrate this method is through a worked example.

Use the Bisection Method to approximate the positive zero of the function $f(x)=x^2-5$`f`(`x`)=`x`2−5 correct to $2$2 decimal places.

Firstly we need to choose an interval over which we think the zero will occur. I'll choose the interval to begin at $x=2$`x`=2 and end at $x=3$`x`=3.

We then substitute each value into the function. You'll know you've chosen a good interval if one value comes back positive and the other negative.

$f(2)=-1$`f`(2)=−1

$f(3)=4$`f`(3)=4

As you can see, my interval was a good choice.

Now we need to bisect our interval, that is, find the $x$`x`-value half-way between $2$2 and $3$3. So $x=2.5$`x`=2.5.

We then substitute that into the function.

$f(2.5)=1.25$`f`(2.5)=1.25

Since this returned a positive value, we now replace our interval point of $3$3 (which also returned a positive value) with $2.5$2.5. You will do this each time, either changing the lower or upper interval value, bringing the interval closer and closer together until you arrive at the approximate zero.

We need a way to do this efficiently so let's create a table.

$a$`a` will represent the lower interval point

$b$`b` will represent the upper interval point

$c$`c` will represent the bisection of the interval

$a$a |
$b$b |
$f(a)$f(a) |
$f(b)$f(b) |
$c$c |
$f(c)$f(c) |
---|---|---|---|---|---|

$2$2 | $3$3 | $-1$−1 | $4$4 | $2.5$2.5 | $1.25$1.25 |

$2$2 | $2.5$2.5 | $-1$−1 | $1.25$1.25 | $2.25$2.25 | $0.0625$0.0625 |

$2$2 | $2.25$2.25 | $-1$−1 | $0.0625$0.0625 | $2.125$2.125 | $-0.484$−0.484 |

$2.125$2.125 | $2.25$2.25 | $-0.484$−0.484 | $0.0625$0.0625 | $2.1875$2.1875 | $-0.215$−0.215 |

$2.1875$2.1875 | $2.25$2.25 | $-0.125$−0.125 | $0.0625$0.0625 | $2.21875$2.21875 | $-0.077$−0.077 |

$2.21875$2.21875 | $2.25$2.25 | $-0.077$−0.077 | $0.0625$0.0625 | $2.234375$2.234375 | $-0.00757$−0.00757 |

$2.234375$2.234375 | $2.25$2.25 | $-0.00757$−0.00757 | $0.0625$0.0625 | $2.24212875$2.24212875 | $0.0274$0.0274 |

$2.234375$2.234375 | $2.2421875$2.2421875 | $-0.00757$−0.00757 | $0.0274$0.0274 | $2.23828125$2.23828125 | $0.0099$0.0099 |

$2.234375$2.234375 | $2.23828125$2.23828125 | $-0.00757$−0.00757 | $0.0099$0.0099 | $2.236328$2.236328 |

What we can see here at the end is that correct to two decimal places, it appears the zero occurs at $x=2.24$`x`=2.24

Consider the functions $h\left(x\right)=3\ln x$`h`(`x`)=3`l``n``x` and $g\left(x\right)=x$`g`(`x`)=`x`. We want to use Newton's method to find an approximation of their point of intersection.

Which function would allow us to use Newton's method to do this?

$f\left(x\right)=3\ln x+x$

`f`(`x`)=3`l``n``x`+`x`A$f\left(x\right)=3\ln x-x$

`f`(`x`)=3`l``n``x`−`x`BConsider the given table of values:

$x$ `x`$1$1 $1.5$1.5 $2$2 $2.5$2.5 $f\left(x\right)$ `f`(`x`)$-1$−1 $-0.28$−0.28 $0.08$0.08 $0.25$0.25 What is the narrowest interval of $x$

`x`in which the x-intercept of $f\left(x\right)$`f`(`x`) lies?Between $x=\editable{}$

`x`= and $x=\editable{}$`x`=Starting with $x_0=1.5$

`x`0=1.5, we will now use one application of Newton's method to find a better approximation to the point of intersection of $h\left(x\right)$`h`(`x`) and $g\left(x\right)$`g`(`x`).First, find the exact value of $f\left(x_0\right)$

`f`(`x`0)Find the exact value of $f'\left(x_0\right)$

`f`′(`x`0)Starting with an initial approximation of $x_0=1.5$

`x`0=1.5, use one application of Newton's method to find a better approximation, $x_1$`x`1, to the point of intersection of $h\left(x\right)$`h`(`x`) and $g\left(x\right)$`g`(`x`). Round your answer to 2 decimal places if necessary.

Consider the equation $x^5-3x+5=0$`x`5−3`x`+5=0. Starting with the initial approximation $x_0=\frac{-3}{2}$`x`0=−32, we want to use one application of Newton's method to find a better approximation to the root of the equation.

Newton's method of approximating roots states:

$x_1=x_0-\frac{f\left(x_0\right)}{f'\left(x_0\right)}$

`x`1=`x`0−`f`(`x`0)`f`′(`x`0)To use this method to find a better approximation to the solution of $x^5-3x+5=0$

`x`5−3`x`+5=0, what must be $f\left(x\right)$`f`(`x`)?$f\left(x\right)=x^5-3x+5$

`f`(`x`)=`x`5−3`x`+5A$x^5-3x=-5$

`x`5−3`x`=−5B$f\left(x\right)=-5$

`f`(`x`)=−5C$f\left(x\right)=x^5-3x$

`f`(`x`)=`x`5−3`x`DFind the exact value of $f\left(x_0\right)$

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

`f`′(`x`0)Starting with an initial approximation of $x_0=\frac{-3}{2}$

`x`0=−32, use one application of Newton's method to find a better approximation, $x_1$`x`1, correct to 2 decimal places if necessary.Here is a graph of $f\left(x\right)$

`f`(`x`).Loading Graph...Fill in the gaps to complete the following statement that explains Newton's method.

Geometrically, Newton's method starts with the initial approximation of $x=\frac{-3}{2}$

`x`=−32, and finds the gradient of the tangent to the curve at that point. This is what $f'\left(\frac{-3}{2}\right)$`f`′(−32) represents. Knowing the gradient of the tangent and a point $\left(\frac{-3}{2},f\left(\frac{-3}{2}\right)\right)$(−32,`f`(−32)) on the tangent line, it then finds the equation of the tangent line. Finally it looks for the $x$`x`-intercept of the tangent line, which gives us our next best approximation $x_1=-1.59$`x`1=−1.59.If we used $x_0=\pm\sqrt[\editable{}]{\editable{}}$

`x`0=±^{}√ as the initial approximation, we would get $f'\left(x_0\right)=\editable{}$`f`′(`x`0)=. The tangent would be parallel to the $x$`x`-axis, and so it would not cut the $x$`x`-axis. Therefore $x_0=\pm\sqrt[\editable{}]{\editable{}}$`x`0=±^{}√ would not be a good initial approximation of the real root.

When a car engine is turned off, its temperature is $71^\circ$71°$C$`C`. It is then left outside in the $\left(-11\right)^\circ$(−11)°$C$`C` cold, and $t$`t` minutes after being turned off, the engine's temperature $T$`T` is given by $T\left(t\right)=82e^{-0.03t}-11$`T`(`t`)=82`e`−0.03`t`−11.

After approximately $t_0=69$`t`0=69 minutes, the engine's temperature reaches $0^\circ$0°C. We want to use Newton's method to find a better approximation for this time.

What equation are we trying to find a better approximation for?

Select all the correct options.

$T\left(t_0\right)=0$

`T`(`t`0)=0A$T\left(t\right)=0$

`T`(`t`)=0B$82e^{-0.03t}-11=0$82

`e`−0.03`t`−11=0CEvaluate $T\left(t_0\right)$

`T`(`t`0), rounding to 2 decimal places.Evaluate $T'\left(t_0\right)$

`T`′(`t`0), rounding to 2 decimal places.Starting with an initial approximation of $t_0=69$

`t`0=69 minutes, and using the rounded values of the previous parts, use one application of Newton's method to find a better approximation, $t_1$`t`1, to the number of minutes it takes for the car engine to reach $0^\circ$0°$C$`C`. Round your answer to 2 decimal places if necessary.