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      <text>2.1.2.1 The argument of a function</text>
    </name>
    <questiontext format="html">
      <text><![CDATA[<h3><span style="color: inherit; font-family: inherit;">The Argument of a Function</span><br></h3>
<p>The input to a function is sometimes called its <span><strong>argument</strong></span> or <b>independent variable</b>. It is frequently necessary to obtain the output from a function if we are given its argument. For example, given the function \(g(t)=3t+2\) we may require the value of the output when the argument is \(4\). We write this as \(g(t=4)\) or more usually and compactly as \(g(4)\). In this case the value of \(g(4)\) is \(3\times 4+2=14\).</p>

<hr>
<h4 class="HELM_example">Example</h4>
<p>Given the function \(f(x)=3x+1\) find</p>

<ol class="HELM_parts_inline">
<li>\(f(2)\)</li>
<li>\(f(-1)\)</li>
<li>\(f(6)\)</li>
</ol>

<h4 class="HELM_solution">Solution</h4>
<ol class="HELM_parts">
  <li><p>The output from the function needs to be found when the argument or input is \(2\). We 
  need to replace \(x\) by \(2\) in the expression for the function. We find \( f(2)=3\times 2 + 1 = 7\).</p>
  </li>
  <li><p>Here the argument is \(-1\). We find \(f(-1)=3\times (-1) +1 = -2\).</p>
  </li>
  <li> <p>\(f(6)=3\times 6+1=19\).</p></li>
</ol>

<p>It is possible to obtain the value of a function when the argument is an algebraic expression. Consider the following Example.</p>

<hr>
<h4 class="HELM_example">Example</h4>
<p>Given the function <span class="math inline">\(y(x)=3x+2\)</span> find</p>

<ol class="HELM_parts_inline">
<li>\(y(t)\)</li>
<li>\(y(2t)\)</li>
<li>\(y(z+2)\)</li>
<li>\(y(5x)\)</li>
</ol>

<h4 class="HELM_solution">Solution</h4>
<p>The rule for this function is ‘multiply the input by \(3\) and then add \(2\)’. We can apply this rule whatever the argument.</p>
<ol class="HELM_parts">
  <li><p>In this case the argument is \(t\). Multiplying this by \(3\) and adding \(2\) we find 
   \(y(t)=3t+2\). Equivalently we can replace \(x\) by \(t\) in the expression for the function, so, \(y(t)=3t+2\).</p>
  </li>
  <li><p>In this case the argument is \(2t\). We need to replace \(x\) by \(2t\) in the expression 
   for the function. So \(y(2t)=3(2t)+2=6t+2\).</p>
  </li>
  <li><p>In this case the argument is \(z+2\). We find \(y(z+2)=3(z+2)+2=3z+8\). It is important 
   to note that \(y(z+2)\) is <span><strong>not</strong></span> \(y\times (z+2)=yz+2y\) but 
   instead reads ‘\(y\) of \((z+2)\)’ where ‘ of’ means ‘take the function of’.</p>
  </li>
  <li><p>Here we have a complication. The argument is \(5x\) and so there appears to be a 
   clash of notation with the original expression for the function. There is no problem if we 
   remember that the rule is to multiply the input by \(3\) and then add \(2\). The input now is \(5x\). 
   So \(y(5x)=3(5x)+2=15x+2\).</p>
  </li>
</ol>]]></text>
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