Show that $x^2-7x+11.76$ has only one zero between $[2.5,4]$ and choose $x_0$ according to $|\alpha-x_0| < 1/M.$
Showing that it only has one zero:
$$ f(2.5)f(4) < 0$$
There's at least one zero.
$$f'(x) = 2x-7 $$ There is a minimum between 2.5 and 4, and the concavity points downwards. This means that $f(2.5)$ happens before the function crosses the x line and $f(4)$ happens after the minimum, but still under the x line. Because it's a polynomial of 2nd degree it can only have 2 zeros. So only one zero must exist in this range.
Is this good? I never know what to say in these kinds of exercises because it's always intuitive. Is there a simpler way to do this?
2nd part:
$$[2.5,3] \\ M = \frac{3*2}{2*2.5*|f'(x)|} = \frac{6}{5|f'(x)|} \\ .5 < \frac{5\max_{[2.5,3]}|f'(x)|}{6} \Leftrightarrow .6 < .8333333...$$
Is this correct?