First case: If the number of real roots $r$ of $f(x)$ is greater than one, then $f'(x)$ has at least $r-1$ real roots. (The limitation "greater than one" is not necessary but the statement is trivial if $r\le 1$.) Given any two roots $a<b$ of $f(x)$, $f$ is continuous and differentiable on $[a,b]$, so by Rolle's theorem $f'(c)=0$ for some $a<c<b$.
There may be more roots of $f'(x)$ than those between roots of $f(x)$, so the only upper bound is the obvious one of $n-1$. Ask if you need examples. It seems to me that if multiplicity is taken into account that the number of real roots of $f'(x)$ has the same parity (even/odd) as the number of real roots of $f(x)$, but I haven't proven it yet. If multiplicity is not taken into account, the parity can be anything.
Second case: If $f(x)$ has degree $n$ and has $n$ real roots, then each consecutive pair of roots of $f(x)$ defines a root of $f'(x)$, which makes $n-1$ roots of $f'(x)$. Since $f'(x)$ is a polynomial of degree $n-1$, this is all possible roots. This continues for all later derivatives, so you are correct: all its derivatives will have all real roots.
Third case: The contrapositive of the second case tells us that if any of its derivatives have any non-real roots, then $f(x)$ also has some non-real roots.
Fourth case: The converse of the third case is not true. For example, $f(x)=x^2+1$ has two non-real roots, but its derivative $f'(x)=2x$ has one real root.
Comment case: You asked, "Suppose $f'(x)$ is a $5$ degree polynomial with $3$ real roots. What are the possible no. of roots that $f(x)$ can have($3,4,5$ etc.?)."
The formulas for $f(x)$ and $f'(x)$ are given in the diagram, where $C$ is a real constant, zero in the graph. You can see that $f'(x)$ is a degree $5$ polynomial with $3$ real roots.
The dashed horizontal lines show the possible number of real roots of $f(x)$ for varying values of $C$. There are $0$ real roots for $C=3$, $1$ real root for $C\approx. 2.638$, $2$ real roots for $C=1$, $3$ real roots for $C\approx -1.757$, and $4$ real roots for $C=-1.9$. My discussion for the first case shows that there cannot be more than $4$ real roots since $f'(x)$ has $3$ real roots.
Now that you know $f’(x)$ has a real root which is $x=0$, you can use it to draw the sign scheme of $f’(x)$. More specifically, $f’(x)>0$ is when $x>0$ and $f’(x)<0$ is when $x<0$. This means $f(x)$ is a monotone increasing function on $(0,\infty)$ and monotone decreasing on $(-\infty,0)$. This means that in the given interval of $x$, the minimum can occur at $x=0$, and maximum either at $x=\ln(7)$ or $x=-\ln(2)$. Checking both these possibilities we conclude that $$\min_{x=0} f(x)=1$$ $$\max_{x=\ln 7} f(x)=\frac{25}{7}$$
Best Answer
THe derivative method is, probably the best one. But here, you may begin with some algebraic trick
REcall that $$11^x + 13^x + 17^x -19^x=0 \iff \left(\frac{11}{19}\right)^x+\left(\frac{13}{19}\right)^x+\left(\frac{17}{19}\right)^x-1=0.$$ Define now $F(x):=\left(\frac{11}{19}\right)^x+\left(\frac{13}{19}\right)^x+\left(\frac{17}{19}\right)^x-1$ wich is continuous and differentiable on $\Bbb R$.
$F(0)=2>0$ and $\lim_{x\to+\infty}F(x)=-1<0$ (so, there is $M\in\Bbb R$ such that $F(M)<0$). An application of Bolzano's Theorem leads to the existence of, at least, one real (positive, in fact) solution.
But, $F'(x)=\left(\frac{11}{19}\right)^x\ln(11/19)+\left(\frac{13}{19}\right)^x\ln(13/19)+\left(\frac{17}{19}\right)^x\ln(17/19)<0$ for all $x\in\Bbb R$, so the solution is unique.