Answer to Question #140052 in Quantitative Methods for Usman

Question

Usin the method of false position. Find the root of equation x^6 - x^4 - x^3 - 1 = 0

Answer 1

$$x^6-x^4-x^3-1=x^4(x^2-1)-(x^3+1)=$$

$$=x^4(x-1)(x+1)-(x+1)(x^2-x+1)=$$

$$=(x-1)(x^5+x^4-x^2+x-1)$$

$x1=1$

Given $f(x)=x^6-x^4-x^3-1$

Find points $a$ and $b$  such that $a<b$  and  $f(a)f(b)<0$

Let $a=1, b=1.5$

$f(a)=f(1)=1^6-1^4-1^3-1=-2$

$f(b)=f(1.5)=1.5^6-1.5^4-1.5^3-1=1.953125$

$$f(a)f(b)=f(1)f(1.5)<0$$

Using iterative Regula-Falsi Formula, the first approximation,

$$x_1=a-\dfrac{b-a}{f(b)-f(a)}f(a)=$$ $$=1-\dfrac{1.5-1}{1.953125+2}(-2)\approx1.252964$$

$f(1.252964)=-1.562404$

$$x_2=1.252964-\dfrac{1.5-1.252964}{1.953125+1.562404}(-1.562404)\approx$$ $$\approx1.362754$$

$f(1.362754)=-0.574795$

$$x_3=1.362754-\dfrac{1.5-1.362754}{1.953125+0.574795}(-0.574795)\approx$$ $$\approx1.393961$$

$f(1.393961)=-0.147639$

$$x_4=1.393961-\dfrac{1.5-1.393961}{1.953125+0.147639}(-0.147639)\approx$$ $$\approx1.401413$$

$f(1.401413)=-0.034199$

$$x_5=1.401413-\dfrac{1.5-1.401413}{1.953125+0.034199}(-0.034199)\approx$$ $$\approx1.403110$$

$f(1.403110)=-0.007729$

$$x_6=1.403110-\dfrac{1.5-1.403110}{1.953125+0.007729}(-0.007729)\approx$$ $$\approx1.403492$$

$f(1.403492)=-0.000390$

$$x_7=1.403492-\dfrac{1.5-1.403492}{1.953125+0.000390}(-0.000390)\approx$$ $$\approx1.403597$$

$f(1.403597)=-0.000087$

$$x_8=1.403597-\dfrac{1.5-1.403597}{1.953125+0.000087}(-0.000087)\approx$$ $$\approx1.403601$$

$f(1.403601)=-0.000020$

$$x_8=1.403601-\dfrac{1.5-1.403601}{1.953125+0.000020}(-0.000020)\approx$$ $$\approx1.403602$$

$f(1.403602)=-0.000002$

If $\varepsilon=0.0001,$ then $|f(x)|<0.0001$

Root of $f(x)$ is $x=1.403602$