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    <p><code>SetDelayed</code> acts as a scoping construction. Arguments are localized if necessary. Any variables that <em>explicitly</em> match the arguments are bound within this scope, others are not.</p> <pre><code>In[78]:= a[x_] := x^2 + b b = x^4; (* the first x^2 is explicitly present and bound to the argument. The x in x^4 present via b is not bound *) In[80]:= a[x] Out[80]= x^2 + x^4 (* this is what you would expect *) In[81]:= a[y] Out[81]= x^4 + y^2 (* surprise *) In[82]:= a[1] Out[82]= 1 + x^4 (* surprise *) </code></pre> <p>So, what you could do is one of two things:</p> <ul> <li>Use <code>Evaluate</code>: <code>functionB2[x_] := Evaluate[model /. fit];</code></li> <li><p>Make dependence of <code>model</code> on x explicit: </p> <p>In[68]:= model2[x_] = 4/Sqrt[3] - a1/(x + b1) - a2/(x + b2)^2 - a3/(x + b3)^4;</p> <p>In[69]:= functionB3[x_] := model2[x] /. fit;</p> <p>In[85]:= functionB3[Sqrt[0.2]]</p> <p>Out[85]= 2.01415</p></li> </ul> <p><strong>Edit</strong> because of question update<br> Because of your definition of functionB2 any argument value yields the same result, as explained above:</p> <pre><code>In[93]:= functionB2[1] Out[93]= 4/Sqrt[3] - 0.335/(0.435 + x)^2 + 0.347/(0.712 + x)^4 - 0.27/(4.29 + x) In[94]:= functionB2["Even a string yields the same ouput"] Out[94]= 4/Sqrt[3] - 0.335/(0.435 + x)^2 + 0.347/(0.712 + x)^4 - 0.27/(4.29 + x) </code></pre> <p>However, this expression contains x's and therefore it can get a numerical value if we provide a numerical value for x:</p> <pre><code>In[95]:= functionB2["Even a string yields the same ouput"] /. x -&gt; 1 Out[95]= 2.13607 </code></pre> <p>Well, this basically what <code>Plot</code> does too. This is why you still get a plot.</p>
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    1. POabout plotting process -- a further question about "A problem in Mathematica 8 with function declaration"
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    1. USSjoerd C. de Vries
    1. USSjoerd C. de Vries
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      1. VTUpMod
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    1. CO@Sjoerd +1 Allow me to welcome you to the 3K family :)
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      1. USDr. belisarius
    2. CO@belisarius Thanks! Shouldn't you remind me of the three things we normally do? ;-)
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      1. USSjoerd C. de Vries
    3. COThanks, Sjoerd. So, the first step mma deals with the plot command is: evaluating the function to be plotted? I previously thought that the fist step when plotting a function is x takes a value, then the function catch this value and returs a result. If this were right, then func = Sin[x]; Plot[func,{x,0,1}] will never yield a figure. Sorry for this stupid misunderstanding. -_-!
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