$$f(x)\ =\ \frac{\left(2x^2-\tan\left(x\right)\right)}{\ln\left(x-1.4\right)}$$
I'm not understanding what "subintervals of amplitude $d = 0{,}4$" means, if I solve this problem by Newton-Raphson method, how would I get started ?
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Is this using the German decimal comma? That is, the interval is $[0.1,3.3]$ subdivided in sub-intervals of length $0.4$? How is the denominator to be evaluated for $x<1.4$? – Lutz Lehmann Jun 06 '19 at 06:34
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Taking the most de-singularized numerator, the (regular) roots of $f$ are also the roots of $$ g(x)=2x^2\cos x-\sin x $$ The case $x=1.4$ needs a separate consideration.
This has the function table at the nodes of the subdivision \begin{array}{c|ccccccccc} x& 0.10 & 0.50 & 0.90 & 1.30 & 1.70 & 2.10 & 2.50 & 2.90 & 3.30\\\hline g(x)&-0.07993 & -0.04063 & 0.22368 & -0.05941 & -1.73639 & -5.31595 & -10.61277 & -16.57077 & -21.34956 \end{array} This has sign changes in $[0.5,0.9]$ and $[0.9,1.3]$, which now can be refined using bisection or any faster converging bracketing method. You could also use Newton to compute the midpoint of the recent bracketing interval, using another midpoint if it does not stay inside the interval.
Lutz Lehmann
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How is this relevant to determine the roots of $f$, considering that both intervals are outside the domain of $f$? Looking at your comment on the OP I see you realised this. – PierreCarre Jun 06 '19 at 12:24
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I assume that the question is a bad reproduction of a meaningful task. If that is not the case, then the real task is primordially something completely different, the recognition of meaningful domains and the rewriting of the task into a sound problem. – Lutz Lehmann Jun 06 '19 at 14:19
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The way the question is currently posed, there are no roots of $f$ in the proposed interval, except possibly $x=1.4$, if we decide to extend $f$ by continuity to that point.
PierreCarre
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