In the accompanying figure, y=f(x) is the graph of a one to one continuous function f. At each point P on the graph of y=2x^2, assume that the areas OAP and OBP are equal. Here PA, PB are the horizontal and vertical segments. Determine the function f.
Now my approach is this:
First area under $y=2x^2$ and $y=x^2$ at a point x is $x^3/3$. Now we need area $OAP$. If we rotate the axes by π/2 radians and reflect the rotated graph about the y axis, we should still preserve the area of the region OAP.
So, first we rotate the axes by π/2 radians and get the equation of $y=2x^2$ and $y=f(x)$ in the rotated system as
$y=\sqrt{\frac{-x}{2}}$ and
$y=f^{-1}(-x)$
Now to reflect the graph about the y-axis, we replace x with -x to get the transformed equations as
$y=\sqrt{\frac{x}{2}}$—(1) and $y=f^{-1}(x)$—(2)
Now we need to find the area under the two curves (1) and (2) from 0 to f(x).
So we have
Area $OAP = \int_{t=0}^{f(x)}{\sqrt{\frac{t}{2}}-f^{-1}(t) dt}$
Since Area $OAP = x^3/3$, upon differentiating we have
$\sqrt{\frac{f(x)}{2}} – (x-f^{-1}(0)) = x^2$.
Thus we have $f(x)= 2(x^2+x-f^{-1}(0))^2$
And from the figure, and the fact that area OPB tends to 0 as x tends to 0, $f^{-1}(0)=0$
Im unable to find the mistake here, and would be grateful to anyone that could point it out.
Best Answer
The area equation should be set up, instead, as
$$A_{OAP}=\frac13 x^3 = \int_{0}^{2x^2}{\left(\sqrt{\frac{t}{2}}-f^{-1}(t) \right)\ dt} $$ which, after differentiation, yields $$x^2=4x \left[x-f^{-1}(2x^2) \right] $$ or $ f^{-1}(2x^2) =\frac34x$. Solve to obtain $f(x)=\frac{32}9x^2$.