Algebra of Real Functions

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Take f of x equals root x, and g of x equals minus root x. Now add them. Root x minus root x is zero, simple. So f plus g is just the zero function, and zero is defined for every real number, right? Hmm, wait. Put in a negative value of x. The root of a negative number? But that is not a real number at all, is it? So where did our zero function go wrong? Today we learn to add, subtract, multiply and divide functions, and more important, to find the domain of the answer. See, in JEE, the operations are easy, and the marks are lost on the domain. So, let's begin.

0:47Where can we combine two functions?

So, we have two real functions. f goes from D one to R, and g goes from D two to R, okay? The double-lined R on screen is the set of real numbers, and D one and D two are the domains, the inputs each function accepts. Now I want to add, subtract, multiply and divide these functions, just like numbers. With me? But first, one question. For which values of x can you even write f of x plus g of x? Only where both values exist, exactly! If x is outside the domain of f, then f of x simply does not exist, so there is nothing to add, right? So we work only on the common part. Capital D equals D one intersection D two, the inputs that both functions accept. Remember this capital D, got it? Everything today happens inside it.

1:40Sum, difference, product, scalar multiple

Now the operations. They are very natural. On capital D, f plus or minus g, at x, equals f of x plus or minus g of x, okay? The little plus-minus sign on screen just means: read it once with plus, and once with minus. So you find both values, and add them or subtract them. That's it. Next, the product. When f and g are written side by side, it means the product function. f g at x equals f of x times g of x. Multiply the two values. Really simple. And alpha f, where alpha is a constant number. Alpha f at x equals alpha times f of x, so every value of f gets multiplied by alpha, okay?

2:29Quotient: one extra condition

Now division. f by g, at x, equals f of x divided by g of x. Looks the same as the others, right? But think for a second. When you divide two numbers, what can go wrong? Yes, exactly! The bottom cannot be zero. So the quotient is defined only where g of x is not equal to zero. The slash through the equals sign means, not equal, okay? So the domain of f by g is every x in capital D, such that g of x is not zero. Look, the curly brackets on screen mean, the set of all; the little symbol like a curved e means, belongs to; and the colon is read as, such that. First the common domain, then throw out the points where g is zero. Two steps. Got it?

3:18The trap: domain first, simplify later

Now, back to our puzzle. f of x is root x, and g of x is minus root x, okay? Add them: root x minus root x, that is zero. The formula is correct, no problem there. So the domain of f plus g is all real numbers, right? Zero is defined everywhere, isn't it? No, no, careful! This is the JEE trap. Go back to the rule. f plus g lives only on D one intersection D two. Root x needs x greater than or equal to zero, so its domain is zero to infinity, with zero included. The square bracket means the end point is included, the round bracket means it is not, and that sideways eight is infinity, okay? Minus root x has the same domain, so the intersection is zero to infinity. So, f plus g has the formula zero, but its domain is only zero to infinity. Not all real numbers! Puzzle solved: a negative x was never allowed in the first place. And here is the habit toppers follow. Find the domain first, using the original f and g. Simplify only after that. Once you simplify, the square roots disappear, and the warning disappears with them. Clear?

4:35Example 1: domains of f+g, fg, f/g for root x, root(4-x)

Right, on to example one. Let f of x be root x, and g of x be the root of four minus x. Find the domains of f plus g, f g, and f by g, and write f g. Pause the video, and try it first. Done? Okay, let's check. Step one, the separate domains. Root x needs x at least zero, so D one is zero to infinity. The root of four minus x needs four minus x at least zero, so x is at most four. So D two is minus infinity to four, with four included, okay? Now the overlap: from zero onwards, and up to four. So D one intersection D two is the closed interval zero to four, both ends included. For f plus g and for f g, nothing more is needed. Both have domain zero to four. Simple as that. And the product: f g at x is root x times the root of four minus x, that is, the root of x into four minus x. And always write the domain with it: zero to four only, okay? Now f by g. Same domain, zero to four, right? Hold on, careful! For a quotient, the bottom must not be zero. The root of four minus x is zero when x equals four. So x cannot be four, okay? So the domain of f by g is zero to four, with zero included and four left out. See the round bracket at four? That one small bracket can cost you the full question in JEE, so look at it carefully, okay?

6:17Example 2: f+g, f-g, fg and f/g for x^2 and 2x+1

Now a slightly harder one, JEE style. Let f of x be x squared, and g of x be two x plus one. Find f plus g, f minus g, f g, and f by g, each with its domain. Pause and try. Hint: think about the domains first. Done? Okay, let's check. f plus g: x squared plus two x plus one. Easy. Now f minus g. x squared minus two x plus one, right? No, no! That is the trap. The minus sign applies to the whole of g. Put g in a bracket: x squared minus, bracket, two x plus one. That gives x squared minus two x minus one. Both terms change sign, okay? Product: x squared times two x plus one, that is two x cubed plus x squared. Clear? Now the domains. Both x squared and two x plus one are defined for every real number. So the sum, the difference and the product all have domain R, all real numbers. And f by g is x squared over two x plus one. Which x must we throw out? Where the bottom is zero, of course! So we need two x plus one not equal to zero. Two x plus one is zero at x equals minus one half. So the domain is R minus the set containing minus one half: every real number except minus one half, okay? And notice, we read this from g itself, before writing anything else. Domain first, always.

7:57Quick revision: intersect domains, remove g(x) = 0

So, quick revision. Combine f and g only on D one intersection D two. For f by g, also remove every x where g of x is zero. And find the domain first, then simplify, never the other way round. Now go and try the practice problems on this page. Domain first, and you will find them easy. All the best!