I was playing with graphing calculator and I was playing with $f(x)=x(\sqrt[x]{x+1}-1) $ to see will it reach infinity ,but to my surprise the function behaves very weird at the large numbers
I am not sure if this is because large numbers makes some eror in the calculator or it because it has infinitely many extremums
one way to answer this question is to find the derivative of this function and make it equal to zero and then solve for $x$ but the problem is I can't solve $\sqrt[x]{x+1}+x(\sqrt[x]{x+1})(\frac{1}{(x)(x+1)}-\frac{\ln(x)}{x^2})=1$
also the calculator go as far as denoting the points which the function has minima or maxima point , this make me believe it is not an eror 
Short answer: You're being bitten by floating-point rounding error.
Long answer: First, let's rewrite the xth root (which typically isn't a standard function in computer math libraries) in terms of the more common natural logarithm and exponential.
$$f(x)=x(\exp(\frac{\ln(x+1)}{x})-1)$$
Let's break this down step-by-step, introducing temporary variables as needed.
$$t_1 := x + 1$$ $$t_2 := \ln(t_1)$$ $$t_3 := t_2 / x$$ $$t_4 := \exp(t_3)$$ $$t_5 := t_4 - 1$$ $$f(x) := x \times t_5$$
Let's plug in a concrete large value of $x$. For example, $x = 10^{6}$ (aka a million). Doing the arithmetic in standard double-precision gives:
$$t_1 := 1000001$$ $$t_2 := 13.815511557963774$$ $$t_3 := 1.3815511557963774 \times 10^{-5}$$ $$t_4 := 1.0000138156069927$$ $$t_5 := 1.3815606992650942 \times 10^{-5}$$ $$f(x) := 13.815606992650942$$
So far, so good. If you evaluate the expression to high precision in WolframAlpha, and then round to the nearest
double, you get $13.81560699258307$. The error is a mere 4.9 parts per trillion.But suppose that you didn't have access to the full 53 bits (≈15.95 decimal digits), and had to round all of the intermediate results to only 7 significant digits.
$$t_1 := 1000001$$ $$t_2 := 13.81551$$ $$t_3 := 1.381551 \times 10^{-5}$$ $$t_4 := 1.000014$$ $$t_5 := 1.4 \times 10^{-5}$$ $$f(x) := 14$$
We do not get the correctly-rounded result of $13.81561$, but instead only the two significant digits of $14$ — an error of 1.33%.
Taking a close look at the numbers, we see that the error is mostly introduced at the step $t_5 := t_4 - 1$. We're taking a number that's slighty greater than 1, and subtracting 1 from it. This is an example of what's called catastrophic cancellation.