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161 lines
6.6 KiB
HTML
161 lines
6.6 KiB
HTML
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<article id="content" role="main">
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<h1 class="center">Inflection Points </h1>
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<p><span class="center">An Inflection Point </span>is where a curve changes from <b>Concave upward</b> to <b>Concave downward</b> (or vice versa)</p>
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<p>So what is concave upward / downward ?</p>
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<table style="border: 0; margin:auto;">
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<tbody>
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<tr>
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<td><b>Concave upward</b> is when the slope increases:</td>
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<td> </td>
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<td><img src="images/concave-upward.svg" alt="concave upward slope increases" height="149" width="234"></td>
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</tr>
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<tr>
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<td><b>Concave downward</b> is when the slope decreases:</td>
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<td> </td>
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<td><img src="images/concave-downward.svg" alt="concave downward slope decreases" height="149" width="234"></td>
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</tr>
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</tbody></table>
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<p>Here are some more examples:</p>
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<p class="center"><img src="images/concave-examples.svg" alt="concave examples" height="157" width="418"></p>
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<p>Learn more at <a href="concave-up-down-convex.html">Concave upward and Concave downward</a>.</p>
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<h2>Finding where ...</h2>
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<p>So our task is to find <b>where</b> a curve goes from concave upward to concave downward (or vice versa).</p>
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<p class="center"><img src="images/inflection-points.svg" alt="inflection points" height="117" width="509"></p>
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<h2>Calculus</h2>
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<p><a href="derivatives-introduction.html">Derivatives</a> help us!</p>
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<p>The derivative of a function gives the slope.</p>
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<p>The <a href="second-derivative.html">second derivative</a> tells us if the slope increases or decreases.</p>
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<ul>
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<li>When the second derivative is <b>positive</b>, the function is <b>concave upward</b>.</li>
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<li>When the second derivative is<b> negative</b>, the function is <b>concave downward</b>.</li>
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</ul>
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<p>And the inflection point is where it goes from <b>concave upward</b> to <b>concave downward</b> (or vice versa).</p>
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<div class="example">
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<h3>Example: y = 5x<sup>3</sup> + 2x<sup>2</sup> − 3x</h3>
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<p style="float:right; margin: 0 0 5px 10px;"><img src="images/5x3-2x2-3x-concave.svg" alt="5x^3 2x^2 3x inflection point" height="243" width="188"></p>
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<p>Let's work out the second derivative:</p>
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<ul>
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<li>The derivative is <b>y' = 15x<sup>2</sup> + 4x − 3</b></li>
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<li>The second derivative is <b>y'' = 30x + 4</b></li>
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</ul>
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<p> </p>
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<p>And <b>30x + 4</b> is negative up to x = −4/30 = −2/15, positive from there onwards. So:</p>
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<div class="so"> f(x) is <b>concave downward</b> up to x = −2/15</div>
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<div class="so">f(x) is <b>concave upward</b> from x = −2/15 on</div>
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<p>And the inflection point is at x = −2/15</p>
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</div>
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<div class="center80">
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<h3>A Quick Refresher on Derivatives</h3>
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<p>In the previous example we took this:</p>
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<p class="center larger">y = 5x<sup>3</sup> + 2x<sup>2</sup> − 3x</p>
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<p>and came up with this derivative:</p>
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<p class="center larger">y' = 15x<sup>2</sup> + 4x − 3</p>
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<p>There are <b>rules</b> you can follow to find derivatives. We used the <a href="derivatives-rules.html">"Power Rule"</a>:</p>
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<ul>
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<li>x<sup>3</sup> has a slope of 3x<sup>2</sup>, so 5x<sup>3</sup> has a slope of 5(3x<sup>2</sup>) = 15x<sup>2</sup></li>
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<li>x<sup>2</sup> has a slope of 2x, so 2x<sup>2</sup> has a slope of 2(2x) = 4x</li>
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<li>The slope of the <b>line</b> 3x is 3</li>
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</ul>
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</div>
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<p>Another example for you:</p>
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<div class="example">
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<h3>Example: y = x<sup>3</sup> − 6x<sup>2</sup> + 12x − 5</h3>
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<p>The derivative is: y' = 3x<sup>2</sup> − 12x + 12</p>
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<p>The second derivative is: y'' = 6x − 12</p>
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<p> </p>
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<p>And 6x − 12 is negative up to x = 2, positive from there onwards. So:</p>
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<div class="so"> f(x) is <b>concave downward</b> up to x = 2</div>
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<div class="so">f(x) is <b>concave upward</b> from x = 2 on</div>
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<p>And the inflection point is at x = 2:</p>
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<p class="center"><img src="images/x3-6x2-12x-5-inflection.svg" alt="x^3 6x^2 12x 5 inflection point" height="199" width="168"></p>
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</div>
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<p> </p>
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<div class="related">
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<a href="index.html">Calculus Index</a>
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