Continuity on an Interval, and Combinations

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. Take f of x equals one over, one minus x. Apply f to it, and then apply f once more. Simplify, and the whole thing collapses to just x! And x is continuous everywhere, right? So is f of f of f of x also continuous everywhere? Hmm. Keep that question in mind; we will settle it today. For that, we need the rules for combining continuous functions. Questions on where a combined function breaks come up often in JEE, so let's begin.

0:38Which functions are continuous?

So, first, the functions you can always trust. This double-struck capital R is the set of all real numbers, and continuous on R means continuous at every real number, okay? Every polynomial, sine x, cos x, e to the power x, modulus of x with the two vertical bars, and the cube root of x, with the small three on the root sign. No breaks, no jumps, anywhere. Clear? Now tell me. What about tan x? Is it continuous everywhere? Not everywhere! Tan x is sine x over cos x, and where cos x is zero, tan x is not even defined. So rational functions, tan x, log x, the natural log written l n, root x, and the inverse trig functions are continuous on their domains: at every point where they are defined. Okay? For tan x, that means every x not equal to two n plus one, times pi over two, where n is an integer. The slashed equals sign means not equal to. So tan x breaks only at the odd multiples of pi over two, and only because it is undefined there. Got it?

1:49Combining continuous functions

Now, the real power of continuity. Suppose f and g are both continuous at a point, small a. Then, as the double arrow says, f plus g, f minus g, f times g, and modulus of f are all continuous at a too. The plus-minus sign just means both, the sum and the difference. With me? So are max of f and g, the bigger of the two, and min of f and g, the smaller of the two. Right? Now the quotient, f over g. What can go wrong here? Yes, exactly! The denominator can be zero. So f over g is continuous at a, provided g of a is not equal to zero. Clear? Quick check from the course. Where is one over, x squared minus five x plus six, discontinuous? Correct! It is a rational function, so it breaks only where the denominator vanishes. And x squared minus five x plus six is zero at x equals two and at x equals three. So two breaks, and it is continuous everywhere else. Okay?

3:00Composition: a function of a function

Next, composition. Suppose g is continuous at a, and f is continuous at g of a, the point where g sends a. Okay? Then f circle g is continuous at a. The small circle means composition: f circle g of x is f of g of x. First g acts, then f. Clear? So, turn it around. Where can f of g of x break? At two kinds of points. One: where the inner function g breaks. Two: where g of x lands on a point where the outer f breaks. Many students check only the second kind. Keep both in mind; both of today's examples need them.

3:47Careful: facts that trip students up

Now a few facts that students often get wrong. Continuous plus discontinuous is always discontinuous. Why? If the sum were continuous, subtract the continuous function from it, and the other one would be continuous too. That cannot happen. With me? So, take two discontinuous functions. Their sum must be discontinuous too, right? No, no, careful! Not always. Two discontinuous functions can add up to a continuous one, and the same can happen with a product, okay? Here is the course's example. Signum x, written s g n x, is one for positive x, minus one for negative x, and zero at zero. So it jumps at zero. Minus signum x jumps at zero too. But add them, and you get zero everywhere: perfectly continuous! And one more. Let f be one when x is greater than or equal to zero, and minus one when x is less than zero. So f jumps at zero. But modulus of f is one everywhere, a constant, and that is continuous. So modulus f can be continuous even when f is not. Got it?

4:56Example 1: breaks of f(f(x)) and f(f(f(x))), f(x) = 1/(1-x)

Now, remember our puzzle from the start? Let's solve it. f of x equals one over, one minus x. Find the points of discontinuity of f of f of x, and of f of f of f of x. Pause the video, and try it first. Done? Okay, let's check. Use our two kinds of breaks. First, the inner f breaks at x equals one, where one minus x is zero. So f of f of x is not defined at x equals one. Okay? Second, the outer f breaks when its input is one, that is, when f of x equals one. One over, one minus x, equals one, gives x equals zero. Now simplify. f of f of x comes out as x minus one, over x. Looking at this formula, only x equals zero is a problem. So f of f of x breaks only at zero, right? No, no, careful! This is the trap. x equals one was thrown out at the very first step, because the inner f is not defined there. Simplifying cannot bring it back. So f of f of x breaks at zero and at one. Got it? And f of f of f of x? Apply f once more, and it simplifies to just x, our puzzle! But only for x not equal to zero or one, because the inner steps fail there. So it is discontinuous, in fact undefined, at zero and at one, just like f of f of x. Puzzle solved!

6:33Example 2: where y = 1/(u2 + u - 2) breaks, u = 1/(x-1)

Now a harder one. u equals one over, x minus one, and y equals one over, u squared plus u minus two. Find all the points where y is discontinuous, as a function of x. Pause the video and try it. Hint: think of the two kinds of breaks. Done? Okay, let's check. Start with the outer function. y breaks where its denominator is zero. u squared plus u minus two factors as u minus one, times u plus two. So u equals one, or u equals minus two. Now convert back to x. u equals one means one over, x minus one, equals one. So x minus one is one, and x equals two. And u equals minus two means one over, x minus one, equals minus two. So x minus one is minus one half, and x equals one half. Two points, two and one half. Done, right? No, no, wait! Remember the two kinds of breaks? The inner function u is not even defined at x equals one, so y is not defined there either. This is the point many students miss. So y is discontinuous at three points: x equals one half, one, and two. Three, not two. Clear?

7:54Quick revision: algebra of continuity and composition rule

So, quick revision. Polynomials, sine, cos, e to the power x and modulus x are continuous on all of R; the rest, on their domains. Sums, differences, products, modulus, max, min and compositions of continuous functions stay continuous, and f over g needs g of a not equal to zero. And for a composite function, check where the inner function breaks, and where it hits a break of the outer one. Now try the practice problems on this page. All the best!