Find the roots of $f(x)$ that are found on the interval $[0{,}1 ; 3{,}3]$ in subintervals of amplitude $d = 0{,}4$

Question

$$f(x)\ =\ \frac{\left(2x^2-\tan\left(x\right)\right)}{\ln\left(x-1.4\right)}$$

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 ?

Answer 1

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.

Answer 2

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.