Maximum and Minimum Values; Range

Transcript

0:00Introduction

Hello everyone! Let's start with one small puzzle. Take x square minus four x plus three, with x only from zero to three. What is its smallest value, and its largest? Most students check just the two ends. Put x equals zero, you get three. Put x equals three, you get zero. So the values run from zero to three. Done, right? Hmm, not so fast. Something is hiding in the middle, and today you will find it. And look, JEE has asked this idea many times. So, let's begin.

0:36Minimum or maximum: the vertex decides

So, take any quadratic, a x square plus b x plus c, and call it f of x. Remember completing the square from the last concept? It becomes a, times x plus b over two a, whole square, minus D over four a. Here D is the discriminant, b square minus four a c. Clear? Now look only at the bracket: x plus b over two a, whole square. Tell me, can this square ever be negative? Never! Yes, a square is always zero or more. So when a is greater than zero, f of x is at least minus D over four a, and it reaches that value when the bracket is zero, at x equals minus b over two a. Remember? That point is the vertex of the parabola, from the last concept. So for a greater than zero: minimum value minus D over four a. And no maximum, because the square keeps growing as x moves away, doesn't it? Alright, now flip it. Let a be less than zero. Same expression, same minus D over four a. So it is still the minimum, right? No, no, careful! Now a times the square is zero or less, so f of x can only come down from minus D over four a. So for a less than zero, it is the maximum value, at the same x, and there is no minimum. Mixing these up is a classic JEE trap. Check the sign of a first. Got it? Now a topper's trick. Minus D over four a is just the value of f at the vertex, x equals minus b over two a. So skip the D formula: find minus b over two a, and put it into f of x. That's it!

2:22On a closed interval [p,q]

Now, here is a twist. Suppose x is stuck in the closed interval p to q. The square brackets mean both ends, p and q, are included, okay? Now, where can the largest and the smallest values sit? Yes, exactly! Only three candidates. The value at p, the value at q, and the value at the vertex. That's it, nothing else to check. But one condition. The vertex counts only if minus b over two a lies inside the interval p to q. If the vertex is outside, ignore it, and only the two ends remain. Clear? Then just compare: the largest candidate is the maximum, the smallest is the minimum. And the trap? Checking only the two ends and forgetting the vertex. Hmm, sounds like our puzzle, right?

3:12Range of a quadratic over a quadratic

Now the big one. y equals a fraction: a one x square plus b one x plus c one, over a two x square plus b two x plus c two. The small ones and twos are subscripts, just labels. Which values can y take? That set is the range. Here is the clever idea: turn the question around. Pick any one value of y. When does it really belong to the range? Exactly, when some real x gives that y! So cross-multiply. You get a quadratic in x, with y inside its coefficients, okay? And x must be real. So the discriminant of this quadratic must be greater than or equal to zero. The double arrow means implies. That inequality in y gives the range. With me? But careful, two checks. First, some value of y can make the x square coefficient zero. Then it is not a quadratic, and the discriminant says nothing. So check that y separately, okay? Second, if some x is forbidden, say one that makes the denominator zero, remove any y that needs that x. Students skip these checks all the time. You won't, right?

4:27Example 1: range of x² - 4x + 3 on [0, 3]

Now, remember our puzzle from the start? Let's finish it. Find the range of x square minus four x plus three, on the closed interval zero to three. Pause the video, and try it first. Done? Okay, let's check. Call it f of x. First, the two ends. f of zero is three. And f of three is nine minus twelve plus three, that is zero. So the smallest value is zero, the largest is three, and the range is zero to three, right? No, no! That was our puzzle answer, and it misses the vertex. Here b is minus four and a is one, so minus b over two a is four over two: two. And two lies inside zero to three, so it counts. f of two is four minus eight plus three: minus one. Look, minus one is smaller than both ends! This is the value that was hiding in the middle. So the smallest value is minus one, and the largest is three. Range: minus one to three, both ends included. Puzzle solved! On an interval, always check the vertex, okay?

5:39Example 2: range of (x² + x + 1)/(x² - x + 1)

Now a harder one, JEE level. Find the range of y equals x square plus x plus one, over x square minus x plus one. Here x belongs to R: the curved E sign means belongs to, and the double-lined R means all real numbers. Pause and try. Hint: cross-multiply first. Done? Okay, let's check. First, the denominator, x square minus x plus one. Here a is greater than zero and the discriminant D is less than zero, so it is positive for every real x. Never zero, so no x is forbidden. Good, one worry less! Now cross-multiply and collect the terms. y minus one, times x square, minus, bracket y plus one, times x, plus y minus one, equals zero. See, a quadratic in x with y in the coefficients, just like our method. For y not equal to one, the x square coefficient is not zero, so the discriminant works. The equals sign with a slash means not equal to. y plus one, whole square, minus four times y minus one, whole square, greater than or equal to zero. Factor this difference of squares: three minus y, times three y minus one, greater than or equal to zero. Now, many students say: a product greater than or equal to zero, so y lies outside the roots. y at most one third, or y at least three, right? No, no, careful! See, the first factor is three minus y, not y minus three, so the sign is flipped. So y lies between the roots: one third is less than or equal to y, which is less than or equal to three. Last check. We assumed y is not one, so test one separately. Put x equals zero: y is one over one, that is one. So one is included; it already sits inside our interval. So the range is one third to three, both ends included. For any real x, this fraction never goes above three, or below one third. Nice and clean!

7:56Quick revision: vertex value -D/4a, interval ends, range

So, quick revision. For a greater than zero, the minimum is minus D over four a, at x equals minus b over two a; for a less than zero, it is the maximum. On a closed interval, compare both ends and the vertex, if it lies inside. And for the range of a fraction: cross-multiply, discriminant greater than or equal to zero, then check. Now go and try the practice problems on this page. All the best!