By first defining f(x) show that each of the following functions has a root on the interval given:
(a) x^3 + 6x^2 = 9x -2 on (0, 0.5)
(b) 1+ x^2 = x^3 on (1,2)
(c) x^2 = sin x on (0.5, 1)
Thank you!
+118723 (a) x^3 + 6x^2 = 9x -2 on (0, 0.5)
$$x^3+6x^2-9x+2=0$$
If I look at the function
$$f(x)=x^3+6x^2-9x+2$$
then when f(x)=0 the original polynomial will equal zero (because they are equal to each other)
$$\\f(0)=0^3+6*0^2-9*0+2=+2\\\\
f(0.5)=0.5^3+6*0.5^2-9*0.5+2\\$$
$${{\mathtt{0.5}}}^{{\mathtt{3}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{6}}{\mathtt{\,\times\,}}{{\mathtt{0.5}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{9}}{\mathtt{\,\times\,}}{\mathtt{0.5}}{\mathtt{\,\small\textbf+\,}}{\mathtt{2}} = {\mathtt{\,-\,}}{\frac{{\mathtt{7}}}{{\mathtt{8}}}} = -{\mathtt{0.875}}$$
So f(0) is positive and f(0.5) is negative so for some value of x between 0 and 0.5 the function will be 0.
hENCE THERE MUST BE A ROOT BETWEEN X=0 AND X=0.5
I am not sure that that is the presentation that you want, but it certainly is the jist of it. :)
You should be able to do the others yourself now 
