No. of real solution of the equation $2^x = x^2$
$\bf{My\; Solution::}$ Let $f(x) = 2^x-x^2$
Now Differentiate both side w. r to $x\;,$ we get $f^{'}(x) = 2^x\ln(2)-2x$
Now Again Differentiate w.r to $x\;,$ we get $f^{''}(x) = 2^x\cdot \ln^2(2)-2.$
Now Again Differentiate w.r to $x\;,$ we get $f^{'''}(x) = 2^x\cdot \ln^3(2)>0\; \forall x\in \mathbb{R}$
Now I did not understand How can I solve It.
Help me
Thanks
That will tell you there are three real roots.
One root ($2$) is obvious and trivially found. So is the other positive root, which is $4$.
The third (negative real root) cannot be exactly determined but it can be approximated by numerical methods.
You can apply the Intermediate Value Theorem to your function $ \ f(x) \ $ , since it is the difference of two continuous functions defined "everywhere". If we take values of this function at some integer values of $ \ x \ $ , we have
$$ f(-1) \ < \ 0 \ , \ f(0) \ > \ 0 \ , \ f(1) \ > \ 0 \ , \ f(2) \ = \ 0 \ , \ f(3) \ < \ 0 \ , \ f(4) \ = \ 0 \ , \ f(5) \ > \ 0 \ \ , $$
which directly locates two real roots of our equation and points to a third lying in the interval $ \ ( -1 \ , \ 0 ) \ $ (that one would need to be found "numerically": it turns out to be about -0.767) .
It remains for us to show that there aren't any more. For $ \ x \ < \ -1 \ $ , we are able to say that there won't be, since $ \ x^2 \ > \ 1 \ $ , but $ \ 0 \ < \ 2^x \ < \ \frac{1}{2} \ $ . For $ \ x \ > \ 4 \ $ , we can look at the derivatives you calculated. Depending on how fundamental an inequality we want to start from, we could take it that
$$ x \ - \ 1 \ > \ \ln x \ \ \Rightarrow \ \ (x-1) \ \ln 2 \ > \ \ln x \ \ \Rightarrow \ \ 2^{x-1} \ > \ x \ \ \Rightarrow \ \ 2^{x-1} \cdot \ln 2 \ > \ x $$
$$ \Rightarrow \ \ f \ '(x) \ = \ 2^x \cdot \ln 2 \ - \ 2x \ > \ 0 \ \ . $$
Also,
$$ 2^{x-1} \ \cdot \ (\ln 2)^2 \ > \ 1 \ \ , $$
so $ \ f \ ''(x) \ > \ 0 \ $ for $ \ x \ > \ 4 \ $ , meaning that $ \ f(x) \ $ is "concave upward". Thus, $ \ f(x) \ $ continues increasing indefinitely, so there are no further real zeroes for the function, and no other roots of our equation.
EDIT (6/6) -- I suppose we could be a bit more thorough about showing that there are no other real roots between the ones found above. In $ \ ( -1 \ , \ 0) \ $ ,
$$ 0 \ < -2x \ < \ 2 \ \ \Rightarrow \ \ \frac{1}{2} \ln 2 \ < \ 2^{-1} \cdot \ln 2 \ - \ 2x \ < \ f \ '(x) \ < \ 2^0 \cdot \ln 2 \ - \ 2x \ < \ \ln 2 \ + \ 2 \ \ . $$
So $ \ f(x) \ $ is strictly increasing on this interval: there is only the one zero at $ \ \approx \ -0.767 \ $ .
Since $ \ 2^x \cdot (\ln 2)^2 \ $ is a positive number multiplying an exponential function, it is strictly increasing over all real numbers. Thus, the second derivative $ \ f \ ''(x) \ = \ 2^x \cdot (\ln 2)^2 \ - \ 2 \ $ has only one real zero at $ \ x \ = \ \log_2 \left[ \frac{2}{(\ln 2)^2} \right] \ = \ 1 \ - \ 2 \ \frac{\ln (\ln 2)}{\ln 2} \ \approx \ 2.0575 \ = \ \alpha \ $ ; this is the single point of inflection for $ \ f \ '(x) \ $ .
Thus, $ \ f(x) \ $ is positive and only "concave downward" on $ \ (0 \ , \ 2) \ $ ; there are then no other zeroes of $ \ f(x) \ $ in this interval. The situation is a little more complicated in $ \ (2 \ , \ 4) \ $ , since the concavity of $ \ f(x) \ $ changes. The function is negative and only "concave upward" on $ \ ( \alpha \ , \ 4) \ $ ; also, $ \ f \ '(2) \ < \ 0 \ $ . The function $ \ f(x) \ $ then should remain negative on $ \ (2 \ , \ 4) \ $ .
The difference of a transcendental function (the exponential function) and an algebraic function (the "power-law" function) make this a bit untidy to analyze. Key features of $ \ f(x) \ $ don't fall at convenient values of $ \ x \ $ , and don't always appear in quite the intervals we might like...
