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 $$\displaystyle f'(x) =  = \left(x^3+\frac{x^2}{2}\right)^2+\frac{1}{4}\left[\left(\sqrt{2}x^2+\sqrt{2}x\right)^2+(x^2)^2+2(x+1)^2+2\right]&gt;0\;\forall x\in \mathbb{R}$$

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 $f'(x) &gt;0$
For $ -1 \le x &lt; 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 $f'(x) &gt;0$
For $x &lt; -1$,  $f'(x)=1 +(x + x^2) +(x^3 + x^4) +( x^5 + x^6)$. Again, each bracketed term is positive so $f'(x) &gt;0$


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.
An odd polynomial has at least one real root is due to the fact that complex roots come in conjugate that is if $\alpha$ is a complex root of a polynomial then $\overline{\alpha}$ is a root too.