Roots of a Function
A root of a function is a value of $x$ for which $f(x)=0$. The graph of $y=f(x)$ will cross the $x$-axis at points corresponding to the roots of the function.
$f(x)=0$ αို αြေαα်α ေαော $x$ αα်αိုးαျားαို $f(x)$ ၏ root αုαေါ်αα်။ $f(x)$ ၏ root αျားαα် $y=f(x)$ αိုαော graph α $x$ αα်αိုးαို αြα်αွားαော αြα်αှα်αျား ($x$-intercepts) αှα့် α‘αူαူαα်αြα ်αα်။
Continuous Function
A function is continuous when its graph is a single unbroken curve.
αေးαားαော domain α‘αွα်း function αα ်αု၏ graph αα် αြα်αောα်αေαြα်းααှိαျှα် ၎α်း function αို continuous αေါ်αα်။
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Location of Roots
α‘αα်αော်αြαါαုံαွα် $f(c)=0$ αြα ်αα်။ αို့αြောα့် $c$ αα် $f(x)=0$ ၏ root αြα ်αα်။
graph α‘α $f(a) > 0$ αြα ်αြီး $f(b) < 0$ αြα ်αα်။
αို့αြောα့် α‘ောα်αါαှα်αα်αျα်αို αောα်αျα်αွဲαိုα်αα်။
If the function $f(x)$ is continuous on the interval $[a, b]$ and $f(a)$ and $f(b)$ have opposite signs, then $f(x)$ has at least one root, $x$, which satisfies $a < x < b$.
αြားαိုα်း $a$ αှα့် $b$ α‘αွα်း $f(x)$ αα် continuous αြα ်αြီး $f(a)$ ၏ αα္ααာαှα့် $f(b)$ ၏ αα္ααာ αα့်αျα်αα်αြα ်αျှα် $f(x)=0$ αြα ်α ေαော $x$ αα်αိုး (root) α‘αα်းαုံးαα ်αု αှိαα်။
Notice
- $f(a)$ ၏ αα္ααာαှα့် $f(b)$ ၏ αα္ααာ αα့်αျα်αα်αြα ်αျှα် root αα ်αုα‘αα်းαုံးαှိαα် αိုαα်αှာ αα ်αုαာαှိαα်αု ααိုαို။ αα ်αုαα်αိုαို၍αα်း αှိαိုα်αα်။
- $f(a)$ ၏ αα္ααာαှα့် $f(b)$ ၏ αα္ααာ αα့်αျα်αα်αြα ်αျှα် root αα ်αုα‘αα်းαုံးαှိαα် αိုαα်αှာ αα္ααာααြောα်းαျှα် root αုံးαααှိαု ααိုαိုαါ။αα ်αုαα်αိုαို၍αα်း αှိαိုα်αα်။
- root ၏ αα်αေαာαို αα့်αှα်းαြα်းαα် root αα်αိုးαို αှာαူαြα်းααုα်αါ။
- α‘αျို့αော αီαျှαြα်းαျားαα် αα်္αျာαိုα်αာ αုα်αုံးαျားαြα့် αြေαှα်းαα် α‘αွα်αα်αဲαော α‘αြေα‘αေαျိုးαှိαα်αα်။ αိုαို့αော α‘αြေα‘αေαွα် root ၏ αα်αေαာαို αα့်αှα်းαိုα်αြီαိုαါα root ααα်αα်းα α်αြီαု αိုαိုα်αα်။
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| $f(a)>0$ αြα ်αြီး $f(b)< 0$ αြα ်αα်။ αα္ααာαြောα်းαα်။ α€αို့αော α‘αြေα‘αေαွα် root α‘αေα‘αွα်αှာ α ααα်းαြα ်αα်။ | $f(a)< 0$ αြα ်αြီး $f(b)< 0$ αြα ်αα်။ αα္ααာααြောα်းαါ။ α€αို့αော α‘αြေα‘αေαွα် root α‘αေα‘αွα်αှာ α ုံ ααα်းαြα ်αα်။ |
To Find Approximate Solution from Root Loaction (Iterative Method)
$2x^2-x-2=0, x< 0$ ၏ approximate solution αို αှာαြα့်αါαα်။
αေးαားαော αီαျှαြα်းαို α‘ောα်αါα‘αိုα်း αြα်αေးαြα့်αα်။
$\begin{aligned} 2x^2-x-2&=0\\\\ 2x^2-x&=2\\\\ x^2-\frac{1}{2}x&=1\\\\ x\left( x-\frac{1}{2}\right)&=1\\\\ x&=\frac{1}{x-\frac{1}{2}}\\\\ \end{aligned}$αို့αြောα့် $2x^2-x-2=0$ ၏ α‘αြေαα် $y=x$ αှα့် $y=\dfrac{1}{x-\frac{1}{2}}$ αို့၏ αြα်αှα်αို αှာαူαြα်းαှα့် α‘αူαူαα်αြα ်αα်။
$F(x)=y=\dfrac{1}{x-\frac{1}{2}}$ αုαα်αှα်αα်αိုαါα ို့။ αိုα‘αါαေးαားαော αီαျှαြα်းαျားαα် $x=F(x)$ αြα ်αာαα်။
Graph α‘α αြα်αှα်αα် $x=-2$ αှα့် $x=0$ αြားαွα် αှိαα်။ αို့αော်αα်αွေ့αွα် αα်αα့်αေαာ၌ αြα်αα်αို αြိုαα် αိαှိαြα်းααှိαောαြောα့် $x=0$ αွα် graph αှα ်αု αြα်αွားαα်αု αα့်αှα်းαါαα်။ $x=0$ αွα် graph αှα ်αု α‘αှα်ααα် αြα်αွား αျှα် $F(0)=0$ αြα ်αα်αါαα်။
$F(0)=\dfrac{1}{0-\frac{1}{2}}=-2$
αို့αြောα့် $x\ne F(x)$ αြα ်αောαြောα့် $x=0$ αွα် graph αှα ်αု ααြα်αြောα်းαိαိုα်αြီး $x=0$ αα် αေαားαော αီαျှαြα်း၏ solution ααုα်αြောα်း αိαိုα်αα်။
$F(0)=-2$ αှ ααှိαော $-2$ αို root α‘αြα
်αူααြီး αα်၍ αှာαြα့်αါαα်။
$F(-2)=\dfrac{1}{-2-\frac{1}{2}}=-0.4, F(-2)\ne-2$ αြα
်αα်။ αို့αြောα့် $x=-2$ αα် αေαားαော
αီαျှαြα်း၏ solution ααုα်αါ။
α‘αα်αါαα်းα‘αိုα်း ααှိαာαော $F(x)$ αို $x_{\text{new}}$ αုαူααြီး αα်αိုးαျားαို αα်αα်αှာαြα့်αါαα်။
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$x=F(x)$ αြα ်ααα်αု αူαααα်αှα်αဲ့αα်αြα ်αာ ααားαါ α‘αျα်α‘αα်αျားα‘α $x=-0.78$(two decimal places) αα် αေးαားαော αီαျှαြα်း၏ root αြα ်αα်αု αိαှိαိုα်αါαα်။
α‘αိုαါα‘αြေα‘αေαို $x=F(x)$ αα် converge αြα ်αα်αု αေါ်αα်။
αူααေးαားαော αီαျှαြα်း $2x^2-x-2=0, x< 0$ αို quadratic formula αြα့် αြα်၍ check αုα်αြα့်αျှα်...
| $\begin{aligned} &\\ x&=\frac{-(-1)-\sqrt{(-1)^2-4(2)(-2)}}{2(2)}\\\\ &=-0.78\ (\text{two decimal places}), (\quad\because x < 0)\\\\ \end{aligned}$ |
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α‘αα်αါα‘αိုα်း αီαျှαြα်းαα ်αု၏ root αို numerical merthod αြα့် αှာαူαိုα်αα်။ αော်αြαဲ့αော αα်းαα်းαို iterative method αုαေါ်αα်။ Iterative method αို α‘αုံးαြုαြီး α‘αြေαှာαြα်း process αို Iteration αုαေါ်αα်။ Iteration αို Engineering, Computer Programming, Business α αော αα်αα်αျားαွα် α‘αုံးαျαα်။
Iterative Method
When trying to solve $f(x)$, rearrange the equation to the form $x=F(x)$. When $\alpha$ is the value of $x$ such that $x=F(x)$ then $\alpha$ is also a solution of $f(x)=0$.
Iterative Formula
$$\begin{array}{|c|} \hline x_{n+1}=F(x_n)\\ \hline \end{array}$$Question (1)
$f(x)=x^{2}-6 x+2$.
(a) Show that $f(x)=0$ can be written as $x=6-\dfrac{2}{x}$.
(b) Starting with $x=4$, use iterative formula to find a root of the
equation $f(x)=0$. Round your answers to 3 decimal places.
(c) Use the quadratic formula to find the roots to the equation $f(x)=0$,
leaving your answer in the form $a \pm \sqrt{b}$,
where $a$ and $b$ are constants to be found.
Question (2)
$f(x)=x^{2}-6 x+1$.
(a) Show that the equation $f(x)=0$ can be written as $x=\sqrt{6 x-1}$.
(b) Sketch on the same axes the graphs of $y=x$ and $y=\sqrt{6 x-1}$.
(c) Write down the number of roots of $f(x)$.
(d) Use your diagram to explain why the iterative formula $x=\sqrt{6 x-1}$
converges to a root of $f(x)$ when $x_{0}=2$.
Question (3)
The equation $e^{x}-1=2 x$ has a root $\alpha=0$.
(a) Show by calculation that this equation also has a root, $\beta$,
such that $1 <\beta < 2$.
(b) Show that this equation can be rearranged as $x=\ln (2 x+1)$.
(c) Use an iteration process based on the equation in part (b),
with a suitable starting value, to find $\beta$ correct to 3 significant
figures. Give the result of each step of the process to 5 significant figures.
Question (4)
(a) By sketching a suitable pair of graphs, show that the equation
$x^{3}+10 x=x+5$ has only one root that lies between 0 and $1 .$
(b) Use the iterative formula $x_{n+1}=\dfrac{5-x_{n}^{3}}{9}$,
with a suitable value for $x_{1}$, to find the value of this
root correct to 4 decimal places. Give the result of each iteration
to 6 decimal places.
Question (5)
The terms of a sequence, defined by the iterative formula
$x_{n+1}=\ln \left(x_{n}^{2}+4\right)$, converge to the value $\alpha$.
The first term of the sequence is $2$ .
(a) Find the value of $\alpha$ correct to $2$ decimal places.
Give each term of the sequence you find to $4$ decimal places.
(b) The value $\alpha$ is a root of an equation of the form $x^{2}=f(x)$.
Find this equation.
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