Limits, Continuity & Differentiability
Differentiability at Junction Points
Grade 12
Question:
<p>Let <i>h</i>(<i>x</i>) = min{<i>x</i>, <i>x</i><sup>2</sup>} for every real number <i>x</i>. Then which of the following is true?</p>
<p>(a) <i>h</i> is not continuous for all <i>x</i></p>
<p>(b) <i>h</i> is differentiable for all <i>x</i></p>
<p>(c) <i>h</i>'(<i>x</i>) = 1 for all <i>x</i></p>
<p>(d) <i>h</i> is not differentiable at two values of <i>x</i></p>
Step-by-Step Solution
Key Concept: At points where the min function switches between two curves, check if the derivatives match to determine differentiability.
<p><strong>Solution:</strong></p><p>The function <i>h</i>(<i>x</i>) = min{<i>x</i>, <i>x</i><sup>2</sup>} can be analyzed by finding the points of intersection of <i>y</i> = <i>x</i> and <i>y</i> = <i>x</i><sup>2</sup>.</p><p>Setting <i>x</i> = <i>x</i><sup>2</sup>: <i>x</i> - <i>x</i><sup>2</sup> = 0, which gives <i>x</i> = 0 or <i>x</i> = 1.</p><p>The function <i>h</i>(<i>x</i>) is continuous everywhere (both <i>x</i> and <i>x</i><sup>2</sup> are continuous).</p><p>However, at the points of intersection <i>x</i> = 0 and <i>x</i> = 1, the function transitions between two different functions (<i>x</i> and <i>x</i><sup>2</sup>), causing changes in the derivative.</p><p>At <i>x</i> = 0: derivative of <i>x</i> is 1, derivative of <i>x</i><sup>2</sup> is 0 (left and right derivatives differ).</p><p>At <i>x</i> = 1: derivative of <i>x</i> is 1, derivative of <i>x</i><sup>2</sup> is 2 (left and right derivatives differ).</p><p>Therefore, <i>h</i> is not differentiable at exactly two values of <i>x</i> (namely <i>x</i> = 0 and <i>x</i> = 1).</p><p>∴ Answer is (d).</p>
Correct Answer: d