Part C: Functions

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Suppose f of x equals a x plus b, a straight line. And when you apply f twice, f of f of x, you get four x plus nine. Find f. Most students say two x plus three, and move on. But wait, there is a second answer hiding here! Today is Part C of our unit problem bank: functions. Mixed problems, just like the exam, and one of them is a real IITJEE question. So, let's begin.

0:36Composition: the inside function acts first

So, most problems in this part put one function inside another. Let's warm up that idea first. g circle f of x, with the small circle you see on screen, means g of f of x, okay? Look at the picture. x starts in set A, on the left. f carries it to f of x, in set B, the middle oval. Only then does g act, and carries it on to g of f of x, in the last oval, set C, okay? So, first f, then g. The inside function always acts first. Clear? Now tell me. If I swap the order and do f circle g, will I get the same answer? Usually not! Composition is not commutative, so the order matters. Don't believe me? Let's check with real numbers.

1:23Order matters: a quick check

Take f of x equals the greatest integer of x, the square brackets on screen, and g of x equals mod x, the two straight bars. And take x equals minus five by three. For g circle f, f acts first. So, what is the greatest integer of minus five by three? It is about minus one point six seven, so the answer is minus one, right? No, no, careful! This is the classic trap. We need the greatest integer that is less than or equal to minus five by three. Minus one is bigger than minus one point six seven, so it does not qualify. Look, on the number line, the greatest integer always sits at x or to its left. Just left of minus one point six seven, the first integer is minus two. So the answer is minus two, not minus one. Got it? Now g acts on it: mod of minus two is two. So g circle f, at minus five by three, gives two. Now the other order, f circle g. This time g acts first: mod of minus five by three is five by three, about one point six seven. And the greatest integer of that is one. Two and one. Different answers from the same two functions! Their difference is one. So always read the order carefully. With me?

2:44A function that is its own inverse

Next, a beautiful one. f of x equals three x plus two, over five x minus three, with x not equal to three by five. Let's put f inside itself and see what happens. Put the whole fraction in place of x, and multiply the top and the bottom by five x minus three. The top becomes nine x plus six, plus ten x minus six: nineteen x. The bottom becomes fifteen x plus ten, minus fifteen x, plus nine: just nineteen. Nineteen x by nineteen. So f of f of x is x itself! Now think. Applying f twice gives back x. So what is f inverse here, the f with a small minus one on top? Yes, f itself! f circle f is the identity function, written capital I, so f inverse equals f. This function undoes itself, like pressing a switch twice. And here is the topper's shortcut. For a x plus b, over c x plus d, check just one thing: is d equal to minus a? Here, a is three and d is minus three. Yes! Self-inverse in one second, no algebra. As simple as that.

4:02Example 1: find every f with f(f(x)) = 4x + 9

Now, remember our puzzle from the start? Let's finish it. f of x equals a x plus b, and f of f of x equals four x plus nine for all x. Find all possible f. Pause the video and try it first. Done? Okay, let's check. Inside first: put the whole of f of x into f. That gives a times, a x plus b, plus b. So, a squared times x, plus a b, plus b. This must equal four x plus nine for every x, so we compare coefficients. The x terms give a squared equals four. The constants give b times a plus one, equals nine, okay? So a is two. Then three b equals nine, b is three, and f of x is two x plus three. Done? Wait. Is two the only number whose square is four? No! Minus two also works. This is exactly the trap from our puzzle. With a equals minus two, the second equation becomes minus b equals nine, so b is minus nine. f of x equals minus two x minus nine. Plug it back in: minus two times minus two x is four x, and the constants give plus nine. It works perfectly! Two answers, not one.

5:27Example 2: where f meets its inverse (IIT-JEE 1995)

Now a harder one, a real IITJEE question from nineteen ninety five. f of x equals, x plus one, whole squared, minus one, for x greater than or equal to minus one. Find all x where f of x equals f inverse of x. Pause the video and try it. Done? Okay. Most students first find f inverse: put y equals f of x, solve for x, square roots everywhere, and then equate. Long road, and easy to slip. So think: do we really need f inverse at all? See, for x greater than or equal to minus one, x plus one is never negative, so as x grows, f grows too. So f is increasing, and its range is minus one to infinity. The square bracket at minus one means minus one is included, and the round bracket at infinity means there is no end. Now remember the trick from chapter four. When f is increasing, the graphs of f and f inverse meet only on the line y equals x. So f of x equals f inverse of x exactly where f of x equals x. No inverse needed! Expand: x plus one, whole squared, minus one, is x squared plus two x. Set it equal to x: x squared plus x equals zero, so x times, x plus one, equals zero. So x is zero, or x is minus one. Always check the domain: both are greater than or equal to minus one, so both are allowed. Answer: the set zero, minus one. Two lines instead of a page! But careful: this trick is only for increasing functions. For a decreasing function, f and f inverse can meet away from the line y equals x, okay?

7:22Quick revision: composition order, self-inverse, f(x) = x

So, quick revision. One: in a composition, the inside function acts first, and the order matters. Two: a x plus b, over c x plus d, is its own inverse when d equals minus a, okay? Three: for an increasing function, solve f of x equals x, instead of f of x equals f inverse of x. Now go to the practice problems on this page. Treat them like a timed test, and write your answer before you read the solution. You can do it. All the best!