Counting Points of Non-differentiability

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle before we start. The absolute value of x has a sharp corner at zero, so it is not differentiable there. You remember that, right? Now take the absolute value of x, and cube it. Same bars, and it is still zero at zero. So, another corner at zero? Surprise! The answer is no. It is perfectly differentiable there. So which points really are corners? Counting such points comes up often in JEE, and today you will learn to do it quickly, without the traps. So, let's begin.

0:38Where does the absolute value get a corner?

So, here is our setting. Take a function g that is differentiable, and put it inside absolute value bars, the two vertical lines on screen. Where g is positive, the bars change nothing. Where g is negative, they flip it upwards. So tell me, where can a corner appear? Yes, exactly! Only where g becomes zero, at a root of g. Say the root is a, so g of a equals zero. If g changes sign there, and g prime of a is not zero, we call it a simple root. The dash after g is called prime, for the derivative, and the slashed equals sign means not equal. Clear? Now, why a corner? On one side of a, the absolute value of g is just g, so its slope is g prime of a. On the other side, it is minus g, so its slope is minus g prime of a. Since g prime of a is not zero, these two slopes are different. Okay? Left-hand derivative not equal to right-hand derivative: that is a corner! So the absolute value of g is not differentiable at a simple root. Right? But now think. What if g prime of a is also zero? Then both slopes are zero, and the two sides agree! No corner, so the absolute value of g is differentiable there. This happens at a root of even multiplicity, like one from a squared factor, or wherever g prime is zero too. So here is the trap: do not mark every root as a corner. Only the simple roots. Got it?

2:16Powers and products: corners can disappear

Now, back to our puzzle. On screen: the absolute value of x, to the power n, where n is greater than one. The sign after n means greater than. Our cube is n equals three. At zero, take a small step h. The slope fraction is the absolute value of h to the power n, minus zero, over h. Its size is the absolute value of h, to the power n minus one. That power is positive, so as h tends to zero, it tends to zero. The arrow means tends to. With me? So the slope is zero from both sides. The absolute value of x, to the power n, is differentiable at zero, when n is greater than one. The power flattens the corner. Puzzle solved! Okay? Next, x times the absolute value of x. The absolute value of x has a corner at zero. So x times it must have a corner too, right? No, no, careful! Check the slope fraction at zero: h times the absolute value of h, minus zero, over h. Cancel h, and we get just the absolute value of h, which tends to zero. Clear? So x times the absolute value of x is differentiable at zero, with slope zero. In fact, the product of a differentiable function and a cornered one can still be differentiable. Right? But a sum is different. Differentiable plus not differentiable is never differentiable, at that point. Why? If the sum were differentiable, subtract the differentiable part, and the other part would become differentiable too. Impossible! So a corner in one term of a sum never disappears. Got it?

3:58Greatest integer, max and min

Two more sources of trouble. First, the square brackets on screen. Square brackets around x mean the greatest integer less than or equal to x; here g of x sits inside. Remember, a differentiable function is always continuous, though the reverse is not true. So a discontinuous function is never differentiable. So the greatest integer of g of x is not differentiable wherever it is discontinuous. Okay? Second, the max or the min of two functions. Max means the larger of the two values, and min the smaller. As x moves along, the max can switch from one function to the other. So, where can a corner appear? Yes, exactly! At the crossing points, where the two graphs meet. With me? At a crossing, the slope jumps from one function's slope to the other's. Usually these two slopes differ, and that gives a corner. So check each crossing point, and compare the two slopes. Clear?

4:59Topper's trick: list, test, count

Now, a trick toppers use for these questions. Step one, list every candidate: the roots inside the bars, the jumps of the greatest integer, and the crossing points of max and min. Okay? Step two, test each candidate. Keep a root only if it is simple, keep a jump only if the whole function is really discontinuous there, and keep a crossing only if the two slopes differ there. Double roots? Out! With me? But one care point. Suppose a factor outside the bars is also zero at that root, like the x in x times the absolute value of x. Then the corner disappears, and the function is differentiable there. So in a product, always check the slopes directly. Clear? Step three, count the distinct points. And when the terms of a sum have corners at different points, those corners can never cancel each other, so each one counts.

5:53Example 1: where |x squared - 3x + 2| is not differentiable

Right? Now, Example one. Where is f of x, equal to the absolute value of x squared minus three x plus two, not differentiable? The small two on x means squared. Pause the video, and try it first. Done? Okay, let's check. Step one, list the candidates. The inside, x squared minus three x plus two, factorises as x minus one, times x minus two. So the roots are one and two. With me? Now step two. Are these roots simple? Let's check with g prime. Here g of x is the inside, so g prime of x is two x minus three. At one, it is minus one. At two, it is plus one. Neither is zero, so yes, both roots are simple. Clear? So both roots give corners. f is not differentiable at x equals one, and at x equals two. Two points, done in three lines! Okay?

6:55Example 2: |(x-1) squared (x-2)|, double root vs simple root

Now a slightly harder one. Is the absolute value of, x minus one, squared, times x minus two, differentiable at one, and at two? Pause the video, and try it. Alright. The inside is zero at one, and at two. So, two roots, two corners, not differentiable at both, right? No, no, careful! First, x minus one, squared, is never negative, so it comes out of the bars as it is. Only x minus two stays inside the absolute value. Okay? Now at one. This root comes from a squared factor, so it is a double root. The slope there is zero, from the left and from the right. With me? So the function is differentiable at x equals one. Just like the cube in our puzzle, the square removes the corner. Right? Now at two. This root is simple. At two, x minus one, squared, equals one, so the slopes are minus one from the left, and one from the right. Different slopes: a corner. Clear? So the function is not differentiable at x equals two. Final answer: at one, yes; at two, no. If you said not differentiable at both, you fell for the trap. Got it?

8:14Quick revision: simple roots, x|x|, [g(x)], max and min

So, quick revision. One: the absolute value of g has corners only at the simple roots of g. Two: double roots give no corner, and both the absolute value of x to the power n, for n greater than one, and x times the absolute value of x, are differentiable at zero. Three: also check where the greatest integer of g of x is discontinuous, and where max and min cross. Now go and try the practice problems on this page. You will count these points in seconds. All the best!