Prove, that two equations are equivalent EDIT: Missed something very important! Sorry! We have $x^4+1=2(2x-1)^{1/4}$ not $x^4+1=2\sqrt{2x-1}$. One friend of mine told me that the equation $x^4+1=2(2x-1)^{1/4}$, where $x\geq \frac{1}{2}$
is equivalent to $$x^4+1=2x$$. How did he obtain this? The equation $x^4+1=2(2x-1)^{1/4}$ is equivalent to $\frac{x^4+1}{2}=\sqrt{2x-1}$ and we can consider two functions: One is        $f:[\frac{1}{2},\infty)\rightarrow[0,\infty)$, $f(x)=(2x-1)^{1/4}$ and                       $g:[0,\infty)\rightarrow [\frac{1}{2},\infty), g(x)=\frac{x^4+1}{2}$. $ g $is the inverse of$ f$ so they can only intersect on the line $y=x$. Then we can consider the equation $x^4+1=2x$
Is that enough? 
 A: What you can say is that $$x^4 +1 = 2\sqrt{2x - 1}\tag{1}$$  $$x^4 + 1 = 2x\tag{2}$$ 
share one solution: when $x = 1$, both equations are satisfied. 
But each equation has a second solution not shared by the other. 


*

*Real solutions to $(1)$ are $x = 1,\;x\approx 0.68682$.

*Real solutions to $(2)$ are $\;x = 1,\;\text{ and }\;x \approx
   0.54369.$
Hence, equations $(1)$ and $(2)$ cannot be equivalent.
Note also that $$2x = 2\sqrt{2x-1} \iff x = 1.$$ The equality holds if and only if $x = 1$, and nowhere else.

We can see each of $f(x) = x^4 + 1, \;g(x) = 2\sqrt{2x - 1},\;h(x) = 2x$ in the following graph (refer to top graph only):

A: This is false. Suppose the equivalence as given: $x^4 + 1 = 2\sqrt{2x-1}$ exactly when $x^4 + 1 = 2x$ for $x \ge 1/2$. In particular, this means that if we assume $x^4 + 1 = 2x$, then 
$$x^4 + 1 = 2\sqrt{2x-1} = 2\sqrt{(x^4 +1)-1} = 2x^2,$$
so $x^4 - 2x^2 + 1 = (x^2 -1)^2 = 0$ and $x = 1$.
But noting the polynomial $p(x) = x^4 - 2x + 1$ has $p(1) = 0$ and dividing by the factor $x - 1$ gives a quotient $q(x) = x^3 + x^2 + x - 1$. Then $q(1/2) < 0$ and $q(1) > 0$ means $q$ has a root between 1/2 and 1, and so therefore does $p$. In other words, $x^4 + 1 = 2x$ and $x \ge 1/2$ does not imply $x = 1$, as we found by taking your friend's assumption.
