A Morse function $f(x)\colon \mathbb{R}^n\to \mathbb{R}$ is a smooth function s.t. all singular points are non-degenerate. A theorem of Sard implies that for any smooth $f(x)$ and almost all $a\in \mathbb{R}^n$ the function $f(x) + ax$ is Morse. 


A Morse function $f$ is called perfect Morse if in addition the singular values are distinct. 


Question: Is there an analog for perfect Morse, i.e., given smooth $f(x)$ is $f(x) + ax$ perfect Morse for almost all $a$? 



Looking for the perfect morse
The answer to your question is negative. A counterexample is given by the Morse function


$f:\mathbb{R}\to \mathbb{R}, \;\; f(x)=\sin(x). $


Using this example it is easy to produce higher dimensional counterexamples.


(The Morse functions with at most one critical point per level set were called *excellent* by R. Thom. They are also known as *stable functions*. They form an open and dense subset in the space of smooth functions on a manifold.)


**Re-Edit** (*Following Mark Grant's observation.*) If $f:\mathbb{R}\to\mathbb{R}$ is a polynomial 


$$f(x)=x^n+\sum\_{j=0}^{n-1} a\_j x^j, $$ 


then your guess is correct. Namely for all but finitely many $a$'s the function $f\_a(x):=f(x)-ax$ is excellent. Here is the argument. 


Consider the real algebraic set $\newcommand{\bR}{\mathbb{R}}$


$$ Z=\lbrace (x,a) \in\bR^2;\;\; f'(x)=a\rbrace. $$


We have a natural projection $\alpha: Z\to \bR$, $(x,a)\mapsto a$, whose fibers are finite sets. This is a semialgebraic map so that there exist finitely many real numbers 


$$A\_0< A\_1< ... < A\_N$$


such that the map $\alpha$ is a $C^3$-covering map over each of the intervals $(-\infty, A\_0)$, $(A\_0,A\_1)$, ... $(A\_N,\infty)$.


Let $I$ be one of these intervals. The set 


$$ Z(I):=\lbrace (x,a)\in Z;\;\;a\in I\rbrace $$


consists of the graphs of finitely many twice differentiable semialgebraic functions 


$$ u\_1,...,u\_m: I\to\mathbb{R} $$


such that $ u\_i(a) < u\_{i+1}(a)$, $\forall a, i$. In other words, the set of critical points of $f\_a$, $a\in I$ is $u\_1(a),\dotsc , u\_m(a)$.


I claim that for all but finitely many $a$'s in $I$ the function $f\_a$ is excellent. I argue by contradiction. If this were not the case, then the semialgebraicity of the $u\_i$'s implies that there exist an open interval $ J\subset I$ and pair of indices $(i,j)$, $i\neq j$, such that


$$ f\_a( u\_i(a) )= f\_a( u\_j(a) ), \;\;f'(u\_i(a))=f'(u\_j(a))=a,\;\;\forall a\in J.$$


The first equality can be rewritten as


$$ f(u\_i)- f(u\_j)= a(u\_i-u\_j). $$


Differentiating the above equality with respect to $a$ and recalling that $f'(u\_i)=f'(u\_j)=a$ we deduce


$$ a(u\_i'-u\_j')= (u\_i-u\_j)+ a(u\_i'-u\_j'). $$


This clearly implies 


$$ u\_i(a)=u\_j(a),\;\;\forall a\in J. $$


*Contradiction!*


**Note.** The same is true for any polynomial $f$ in any number of variables. The [proof in the general case](http://liviusmathblog.blogspot.com/2012/07/excellent-polynomial-mappings.html) is very similar to the one above in the one-dimensional case.


