Domain of Real Functions

Transcript

0:00Introduction

Hello everyone! Let's start with one small experiment. Take f of x equals root of x minus one, plus root of five minus x. Put x equals one. Root zero plus root four: two. No problem. Now put x equals ten. Root nine is three, fine. But root of minus five? Oh no, the calculator just says error! So which values of x is this function allowed to take? That set is the domain, and today you will learn to find it for any formula. And look, we will even crack a JEE Main two thousand nineteen question. So, let's begin.

0:40Who decides the domain?

So, first, what is a real function? It is a function whose domain and codomain are both subsets of the real numbers, the double struck R you see on screen, okay? But often, a question gives you only a formula, like our f of x. Nobody tells you the domain. So who decides it? Yes, the formula itself decides! The domain is the largest set of real x for which the formula gives a real number. So, throw out every x that gives an error, and keep the rest. Clear?

1:12One condition for each dangerous part

Look, for every dangerous part of a formula, we write one condition. First, a fraction: one over g of x. You can never divide by zero, right? So, g of x is not equal to zero, the equals sign with a slash. Second, the square root, or any even root: g of x is greater than or equal to zero. Zero is allowed, because root zero is just zero. Now one over root g of x. Same condition, right? No, no, careful! If g of x is zero, root g is zero, and that zero sits in the denominator. So here g of x must be strictly greater than zero. This is a classic JEE trap. Don't fall for it! Next, log to the base a, of g of x. Here also, g of x must be strictly greater than zero. Zero is not allowed, negative is not allowed, okay? And if the base itself has an x in it, log to the base h of x, of g of x, then three conditions. g of x greater than zero, h of x greater than zero, and h of x not equal to one. Students forget that last one again and again. Got it?

2:27More conditions, and the golden rule

Alright, a few more. Sine inverse and cos inverse of g of x; the small minus one on top means inverse, okay? Now tell me, sine or cos of any angle always lies between which two numbers? Minus one and one, exactly! So here g of x must lie between minus one and one, both ends included. Then n C r and n P r. Can you choose two and a half objects? Or minus two objects? No, right? So n and r must be integers, with r from zero up to n, both included, okay? And now the friendly ones: cube root, modulus, greatest integer, and a to the power g of x, with a positive. No extra condition at all. Just relax. Now, if a formula has two parts, like our two roots, do we take the union of the conditions, or the intersection? Intersection, yes! Both parts must work for the same x. So the domain is the intersection of all the conditions, the common part. Not the union. Clear?

3:38Topper's trick: the wavy curve method

Now let me share a trick that toppers use. Very often, your conditions become an inequality like this: x minus a one, times x minus a two, and so on, over x minus b one, and so on, greater than or equal to zero. The small numbers below a and b just label the roots. To solve it fast, use the wavy curve method. Step one: mark all the roots on the number line, from the numerator and from the denominator. Step two: start with a plus sign at the far right, and then alternate, plus, minus, plus, as you cross each root. The signs go up and down like a wave. Really simple. Step three, one small exception. If a root is repeated an even number of times, the sign does not change across it, okay? And step four: for greater than or equal to, include the numerator roots, because there the value is zero. Denominator roots? Never, you cannot divide by zero. We use the wavy curve in both examples today.

4:42Careful: log inequalities

One warning before the examples. If the base a is greater than one, then log to the base a of t, greater than or equal to k, means t is greater than or equal to a to the power k, okay? The double headed arrow means if and only if. But if the base is between zero and one? Same thing? No, the sign flips! Now it means t is less than or equal to a to the power k. And never forget: t must also be greater than zero, always.

5:15Example 1: domains with square roots by wavy curve

Let's do example one, and part one is our function from the start! Find the domain of root of x minus one, plus root of five minus x. Part two: the root of x minus one, over x minus five. Pause the video, and try it first. Done? Okay, let's check. Part one has two square roots, so two conditions. First, x minus one is at least zero, so x is at least one. Then, five minus x is at least zero, so x is at most five. Intersection: one to five, both ends included; that is what square brackets mean. See? That is why x equals ten was outside the domain! Part two. The whole fraction sits under a root, so it must be greater than or equal to zero. And x minus five is a denominator, so x is not equal to five. Now the wavy curve. The roots are one and five. Far right, from five to infinity: plus. The sideways eight on screen is infinity, okay? Between one and five: minus. Left of one: plus again. Now the end points. Do we include one? Do we include five? At x equals one, the fraction is zero over minus four, which is zero, and root zero is fine. So include one. At x equals five, the denominator becomes zero. Never include it! That is the trap here. So the domain is minus infinity to one, with one included, union five to infinity. Round bracket means not included, and the cup symbol means union. Got it?

6:58Example 2: JEE Main 2019 domain with a log and fraction

Now a harder one, from JEE Main two thousand nineteen. Find the domain of f of x equals one over four minus x squared, plus log to the base ten of x cubed minus x. Pause the video, and try it. Done? Okay, let's check. Start with the log: its argument must be strictly positive, so x cubed minus x is greater than zero. Factorise it: x, times x minus one, times x plus one. The roots are minus one, zero and one. Wavy curve, starting from the right. Plus beyond one, minus between zero and one, plus between minus one and zero, and minus left of minus one. We need greater than zero, so take the plus parts, with no end points. x belongs to minus one to zero, union one to infinity; the curly E means belongs to. So that is the answer, right? No, no, wait! What about the first part? The first part is a fraction, so four minus x squared is not zero: x is not plus or minus two. Is minus two in our set? No, it is already outside. But two is inside, so only two is removed. That's it! The domain is minus one to zero, union one to two, union two to infinity. One condition from each part, then the intersection. That is the whole method. See? That is a JEE Main question, done in a few lines. You can do this!

8:34Quick revision: conditions, wavy curve, intersection

So, quick revision. First, write one condition for every part of the formula. Second, solve with the wavy curve, and check every end point. Third, the domain is the intersection of all the conditions. Now go and try the practice problems on this page. You will find them easy, I promise. All the best!