Let $f(x)=(x+3)^2+\cfrac{9}{4}$ for $x\ge -3 $.Compute the shortest possible distance between a point on the graph of $f$ and a point on the graph of $f^{-1}$.
My effort
Let $P,Q$ be points on the graph of $f$ and $f^{-1}$ ,respectively,such that $PQ$ is the shortest distance possible.
So $P$ has coordinates $(a,f(a))$ and $Q$ $(f(a),a)$ (I've already proved this to be the case)
So $$PQ=\sqrt{(a-f(a))^2+(f(a)-a)^2}=\sqrt{2(a-f(a))^2}=\sqrt{2}|(a-f(a))|$$
Since $y=a-f(a)=-(a+5/2)^2-5$ is always negative we must have that $$y=|a-f(a)|=(a+5/2)^2+5$$ which means that the shortest possible distance occurs when $a=-\cfrac{5}{2}$,so the distance is $5\sqrt{2}$
Is my solution correct?Are there other ways to approach the problem ?
Edit : Correction of an error in the computation of the derivative.
There is at least another approach, a little more general.
Graphs of $f$ and $f^{-1}$ resp. are known to be symmetrical with respect to straight line $B$ with equation $y=x$. Let us grow a strip symmetrically on both sides of $B$ until this strip touches the two graphs. As function $f$ is differentiable, the gradients at the contact points should be the same. Said in an equivalent way, the tangent lines have to coincide. Thus it suffices to impose $f'(x)=1$ (in very close similarity with your condition).
This gives $2(x+3)=1$ ; thus $x=-5/2$ for the abscissa of the contact point.
Remark: In the general case, $f'(x)=1$ is only a necessary condition. It can give spurious solutions.