no. of real roots of the equation $ 1+\frac{x}{1}+\frac{x^2}{2}+\frac{x^3}{3}+............+\frac{x^7}{7} = 0$ 
The no. of real roots of the equation $\displaystyle 1+\frac{x}{1}+\frac{x^2}{2}+\frac{x^3}{3}+............+\frac{x^7}{7} = 0 $

$\bf{My\; Try::}$ First we will find nature of graph of function $\displaystyle 1+\frac{x}{1}+\frac{x^2}{2}+\frac{x^3}{3}+............+\frac{x^7}{7}$
So $$\displaystyle f(x) = 1+\frac{x}{1}+\frac{x^2}{2}+\frac{x^3}{3}+............+\frac{x^7}{7}.$$ Then Differentiate
both side w . r to $x\;,$ We get $$\displaystyle f'(x)=1+x+x^2+x^3+..........+x^6$$
Now for max. and Minimum Put $$f'(x) = 0\Rightarrow 1+x+x^2+x^3+x^4+x^5+x^6 = 0$$
We can write $f'(x)$ as  $$\displaystyle \left(x^3+\frac{x^2}{2}\right)^2+\frac{3}{4}x^4+x^3+x^2+x+1$$
So $$\displaystyle f'(x) = \left(x^3+\frac{x^2}{2}\right)^2+\frac{1}{4}\left[3x^4+4x^3+4x^2+4x+4\right]$$
So $$\displaystyle f'(x) =  = \left(x^3+\frac{x^2}{2}\right)^2+\frac{1}{4}\left[\left(\sqrt{2}x^2+\sqrt{2}x\right)^2+(x^2)^2+2(x+1)^2+2\right]>0\;\forall x\in \mathbb{R}$$
So $f'(x) = 0$ does not have any real roots. So Using $\bf{LMVT}$ $f(x) = 0$ has at most one root.
In fact $f(x) = 0$ has exactly one root bcz $f(x)$ is of odd degree polynomial and it
will Cross $\bf{X-}$ axis at least one time.
My question is can we solve it any other way, i. e without using Derivative test.
Help me , Thanks

$f(x) = 1+\frac{x}{1}+\frac{x^2}{2}+\frac{x^3}{3}+............+\frac{x^7}{7}.$ So, as you note, 
$f'(x)=1+x+x^2+x^3+..........+x^6$. Now consider $x$ in three ranges


*

*For $x \ge 0$ then clearly $f'(x) >0$

*For $ -1 \le x < 0$, $f'(x)=(1+x) + (x^2+x^3) + (x^4+x^5) + x^6$. Each bracketed term is non-negative, and $x^6$ is positive, so $f'(x) >0$

*For $x < -1$,  $f'(x)=1 +(x + x^2) +(x^3 + x^4) +( x^5 + x^6)$. Again, each bracketed term is positive so $f'(x) >0$


From that conclude that $f(x) $ is a monotonic increasing function and can therefore have at most 1 zero. 
Finally, as $x \to -\infty$, $f(x) \to -\infty$ and as $x \to +\infty$, $f(x) \to +\infty$ so that by continuity $f(x) $ must have at least one zero, and from the previous it therefore has exactly one zero. 
