Finding the roots or interval in which the (roots) lie of the given equation.

Question

If an equation $$\frac{1}{x-3} + \frac{1}{x-4} + \frac{1}{x-5}+ \frac{1}{x-7}=0 $$ has been given to us then how can we find out-- 1) the number of roots of the equation ??...if it was one of those easy polynomials I could have easily done it..but here it is different..also

2)can we find out what the roots are with only this much information[ any hint would suffice in this part ..no need to solve it ..fully...that's my job!!]..or even if this can't be done

3)can we make an approximate graph{not an exact one[just approximate]} so that we can at least tell in which interval the roots lie..and also tell what the interval is??

Edit: I have already done this ( removing the denominator and assuming that $x\neq 3,4,5,7$ otherwise denominator will become equal to zero which is not allowed) so here is the new equation I got.. $$(x-4)(x-5)(x-7)+(x-5)(x-7)(x-3)+(x-5)(x-4)(x-3)+(x-7)(x-4)(x-3)=0$$

But what to do now..how to get the things I mentioned above..in the question..

Answer 1

The polynomial is the first derivative of $f(x)=(x-3)(x-4)(x-5)(x-7).$ By inspection, we have $f(x)>0$ for $x<3$, for $x\in (4,5)$,and for $x>7$. And $f(x)<0$ for $x\in (3,4)$ and for $x\in (5,7).$ So $f(x)$ has a negative-valued local minimum in $(3,4)$ ,and also in $(5,7)$,and a positive-valued local maximum in $(4,5)$.These are the $3$ points where $f'=0.$

Answer 2

Hint: Multiply with right terms and remove denominator to get a polynomial of degree three. You can do the rest!

Answer 3

you dont want to multiply out. let $f$ be defined by $f(x) = \frac{1}{x-3} + \frac{1}{x-4} + \frac{1}{x-5}+ \frac{1}{x-7}.$ then we have the following:

(a) $f = \frac 4x +\cdots \text{ for } x \to \pm \infty$

(b) $\lim_{x \to k-} f(x) = -\infty, \lim_{x \to k+} f(x) = \infty, \text{ for }k = 3, 4, 5, 7$

the graph is monotone decreasing in every one of the intervals $(-\infty, 3), (3, 4), \cdots , (7, \infty)$

therefore by the intermediate values theorem, $f$ has $3$ simple roots one in each open intervals $(3, 4), (4, 5), 5, 7).$