The question is to find number of real solutions of $F(x)=x^3+1=2\sqrt[3]{2x-1}=G(x)$
So first i tried to find the roots of it's derivative , which was not helpful as the equation formed was hard to solve. Then i tried to visualize the graph which was again not very helpful.
Luckily i found the above equation is true for $x=1$. At $x=0$, $G(x)$ is under $F(x)$ but it's derivative is greater than $F(x)$ till $x=1$ (easy to show). And at $x=1$ also derivative of $G(x)$ is greater than that of $F(x)$, so here $G(x)$ overtakes (not before $x=1$) $F(x)$.
Now after $x=1$, once the $F'(x)$ becomes greater than $G(x)$, it would remain like that so $F(x)$ once again over take $G(x)$ and hence we get 2 solutions of for $x>0$.
Then again when we go from $x=0$ to negative $x$-axis , we will see once $F(x)$ starts decreasing more rapidly than $G(x)$ it would remain that way, so we will get one solution for $x<0$.
So i got the correct answer of $3$. But this is a lucky solution. I would be happy if you can provide me with any type of solution. There should be a better way to solve this problem right?
Best Answer
HINT.-Taking equivalently $\dfrac{x^3+1}{2}=\sqrt[3]{2x-1}$ it is verified that if you made $y=x$ you do have in both sides the same cubic equation; in fact: $$\dfrac{x^3+1}{2}=x\iff x^3-2x+1=0\\\sqrt[3]{2x-1}=x\iff x^3-2x+1=0$$ so the two curves have its common points with the diagonal $y=x$.
Consequently because it is clear that $(1,1)$ is one of this points the two others are the roots of the quadratic equation $x^2+x-1=0$.
Thus the roots are $x=1$ and $x=\dfrac{-1+\sqrt5}{2}$