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    <p>First of all, remember how <code>fmap</code> is defined for applicatives:</p> <pre><code>fmap f x = pure f &lt;*&gt; x </code></pre> <p>This means that your example is the same as <code>(fmap (+ 5) (* 3)) 4</code>. The <code>fmap</code> function for functions is just composition, so your exact expression is the same as <code>((+ 5) . (* 3)) 4</code>.</p> <p>Now, let's think about why the instance is written the way it is. What <code>&lt;*&gt;</code> does is essentially apply a function in the functor to a value in the functor. Specializing to <code>(-&gt;) r</code>, this means it applies a function returned by a function from <code>r</code> to a value returned by a function from <code>r</code>. A function that returns a function is just a function of two arguments. So the real question is this: how would you apply a function of two arguments (<code>r</code> and <code>a</code>, returning <code>b</code>) to a value <code>a</code> returned by a function from <code>r</code>?</p> <p>The first thing to note is that you have to return a value of type <code>(-&gt;) r</code> which means the result also has to be a function from <code>r</code>. For reference, here is the <code>&lt;*&gt;</code> function:</p> <pre><code>f &lt;*&gt; g = \x -&gt; f x (g x) </code></pre> <p>Since we want to return a function taking a value of type <code>r</code>, <code>x :: r</code>. The function we return has to have a type <code>r -&gt; b</code>. How can we get a value of type <code>b</code>? Well, we have a function <code>f :: r -&gt; a -&gt; b</code>. Since <code>r</code> is going to be the argument of the result function, we get that for free. So now we have a function from <code>a -&gt; b</code>. So, as long as we have some value of type <code>a</code>, we can get a value of type <code>b</code>. But how do we get a value of type <code>a</code>? Well, we have another function <code>g :: r -&gt; a</code>. So we can take our value of type <code>r</code> (the parameter <code>x</code>) and use it to get a value of type <code>a</code>. </p> <p>So the final idea is simple: we use the parameter to first get a value of type <code>a</code> by plugging it into <code>g</code>. The parameter has type <code>r</code>, <code>g</code> has type <code>r -&gt; a</code>, so we have an <code>a</code>. Then, we plug both the parameter and the new value into <code>f</code>. We need both because <code>f</code> has a type <code>r -&gt; a -&gt; b</code>. Once we plug both an <code>r</code> and an <code>a</code> in, we have a <code>b1</code>. Since the parameter is in a lambda, the result has a type <code>r -&gt; b</code>, which is what we want.</p>
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    1. COWow, flashback to the abstract algebra class that cleaned my clock ten years ago! I really do appreciate the patient detail of this response. You're a gifted teacher, and I hope you'll at least consider academia among your post-graduate options.
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