Just solve $$\frac=0,$$ there is no need for calculus.
answered Dec 11, 2017 at 12:09 JP McCarthy JP McCarthy 7,831 1 1 gold badge 35 35 silver badges 58 58 bronze badges $\begingroup$You have to solve $$\frac=0$$to solve it you can just multiply by $x-1$ both sides. $$x^2-3=0$$ hence the 2 solutions are $\pm\sqrt 3$. We know that there are no other solution because $\frac1=0$ has no solutions and that $x^2-3$ is second degree thus it has at most $2$ roots
answered Dec 11, 2017 at 12:40 10k 2 2 gold badges 17 17 silver badges 42 42 bronze badges $\begingroup$$[-2,-1] $ is one such interval since $f(-2)f(-1)= - \frac \cdot 1 \lt 0$
$[1.1,2]$ is another interval in which f has the same properties.
Since both intervals are subsets of$ [-2,2]$, both have a root (by IVT), and their intersection is null, f must have two distinct roots in$ [-2,2]$.
Edit: must have AT LEAST two distinct roots. There could be more (not in this particular case, of course)
answered Dec 11, 2017 at 12:43 Francisco José Letterio Francisco José Letterio 2,146 13 13 silver badges 25 25 bronze badges$\begingroup$ Of course since the exercise is easy in this case you can just find the roots. This is for proving their existence in case finding them is not so easy $\endgroup$
Commented Dec 11, 2017 at 12:49 $\begingroup$$Y’$ is alway greater than $0$ in the domain thus the function is always increasing in the domain at approaching $1$ from the negative side it is equal to infinity for the positive side it is equal to negative infinity and $f(-2)$ is smaller then $0$ so that is one real solution because the function is increasing so it goes from some thing negative to infinity it will pass by $0$ and from negative infinity to posit if something it will go by $0$ too. thus there is two real solutions
15.7k 6 6 gold badges 36 36 silver badges 118 118 bronze badges answered May 2, 2019 at 22:48 user670663 user670663To subscribe to this RSS feed, copy and paste this URL into your RSS reader.
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