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Quadratic Equation Problem.

  1. Aug 12, 2013 #1
    1) Find the range of values of k such that the function [tex]f(x) \equiv k(x+2)^2-(x-1)(x-2)[/tex] never exceeds 12.5. I've missed several stages of the computation because it is quite lengthy. I hope you get the flow of things.

    My attempt...
    [tex](k-1)\left[ x + \left( \frac{4k+3}{k-1} \right)x + \frac{4k-2}{k-1} \right] \le 12.5[/tex]Complete the square on LHS...
    [tex](k-1)\left[ \left( x + \frac{4k+3}{2(k-1)} \right)^2 - \frac{48k+1}{4(k-1)^2} \right] \le 12.5[/tex]This boils down to...[tex]-196k \le -98[/tex] and so [tex]k \ge 0.5[/tex]Solution given is: [tex]k \le 0.5[/tex] not sure why. Any help thanks.
     
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  3. Aug 12, 2013 #2

    arildno

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    First off:
    Can you see what info you may get from the second derivative of f?
    You should get that k must be strictly less than 1, so that your (k-1) factor is necessarily negative.
    That might be the reason why you get the wrong equality sign.
     
  4. Aug 12, 2013 #3
    Sorry, I can't use the calculus technique for this. Yes, i know from the properties of quadratic functions that if the coefficient of x^2 is < 0 then f(x) is a max at x = -b/2a.

    Thanks for the reply.
     
  5. Aug 12, 2013 #4
    I'm not quite sure what you are doing. Also, if your goal is to solve for k, I'm not sure why you have x in your equations.

    -Your first equation is quadratic in x
    -Your second equation is linear in x
    -Your third equation is again quadratic in x
    -The remainder of your equations don't have x

    -Junaid
     
  6. Aug 13, 2013 #5

    Ray Vickson

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    If k > 1 the graph of y = f(x) is a parabola that opens upwards, so cannot remain <= 12.5. If k = 1 the function f(x) is linear in x but not constant, so again cannot remain <= 12.5 always. Therefore, we need ---at least--- k < 1 for the graph of y = f(x) to remain bounded from above (so the graph is a parabola opening downward). Now look at the three cases: (1) f(x) = 0 has two roots; (2) f(x) = 0 has a single root (repeated); and (3) f(x) = 0 has no real roots. In case (1) the maximum of f(x) occurs half-way between the two roots (obvious, but you may want to prove it---without calculus); so you can find the maximum value of f(x) in this case, and require it to be <= 12.5. This gives a range of k, and you can check that as k decreases, the value of max f(x) decreases also. When k decreases far enough, we reach cases (2) and (3) and still have max f < 12.5..
     
  7. Aug 13, 2013 #6
    Ray, went down this route and got exactly the same inequality as before. I need to show this result algebraically. I've tried a number of different routes on this and I always get the same result. The only conculsion that I can draw is that my initial statement[tex]f(x) \equiv k(x+2)^2 - (x-1)(x-2) \le 12.5[/tex]from the given information is wrong. But I can't workout why.

    @Junaid: You've spotted a typo. Thanks.
     
  8. Aug 13, 2013 #7

    Ray Vickson

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    For the problem exactly as written, the solution is that ##f_{\max} \leq 12.5## if and only if ##k \leq 1/2##. The point is that k needs to be smaller than something in order to ensure that the parabola y = f(x) opens downward and so remains bounded from above! All I can suggest is that you do it again---carefully--and if you are still confused, show us the work in detail.
     
  9. Aug 13, 2013 #8
    Ray, been through the calculation, same result. Here is my working in full...
    [tex]f(x) \equiv k(x+2)^2-(x-1)(x-2) \le 12.5[/tex][tex]k(x^2+4x+4)-(x^2-3x+2) \le 12.5[/tex][tex]kx^2+4kx+4k-x^2+3x-2 \le 12.5[/tex][tex](k-1)x^2+(4kx+3)x+4k-2 \le 12.5[/tex]for coefficient of [tex]x^2 < 0, k < 1[/tex] note: not assuming this is the correct result but just an educated guess as to the possible range for k.[tex]\Rightarrow 2(k-1)x^2+2(4kx+3)x+8k-4 \le 25[/tex][tex]2(k-1)\left[x^2 + \left(\frac{4k+3}{k-1}\right)x + \left( \frac{4k-2}{k-1}\right) \right] \le 25[/tex][tex]2(k-1)\left[x^2 + \left(\frac{4k+3}{k-1}\right)x + \left( \frac{4k+3}{2(k-1)} \right)^2 + \left( \frac{4k-2}{k-1}\right) - \left( \frac{4k+3}{2(k-1)} \right)^2 \right] \le 25[/tex][tex]2(k-1)\left[x^2 + \left(\frac{4k+3}{k-1}\right)x + \left( \frac{4k+3}{2(k-1)} \right)^2 + \left( \frac{4k-2}{k-1}\right) - \frac{(4k+3)^2}{4(k-1)^2} \right] \le 25[/tex][tex]2(k-1)\left(x+ \frac{4k+3}{2(k-1)}\right)^2 + 8k-4 - \frac{(4k+3)^2}{2(k-1)} \le 25[/tex][tex]2(k-1)\left(x+ \frac{4k+3}{2(k-1)}\right)^2 + \frac{2(k-1)(8k-4)-(4k+3)^2}{2(k-1)} \le 25[/tex]there is a max @ [tex]x=-\frac{4k+3}{2(k-1)}[/tex] therefore[tex]\frac{2(k-1)(8k-4)-(4k+3)^2}{2(k-1)} \le 25[/tex][tex]\Rightarrow 2(8k^2-12k+4)-(16k^2+24k+9) \le 50(k-1)[/tex][tex]16k^2-24k+8-16k^2-24k-9 \le 50k-50[/tex][tex]-48k-1 \le 50k-50[/tex], therefore: [tex]-98k \le -49[/tex] divide both sides by -98 and [tex]k \ge 0.5[/tex]
    Note that the work here is alittle different as I attempted the problem again before posting but the result is the same. Hope this helps.
     
  10. Aug 13, 2013 #9

    Ray Vickson

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    You are NOT doing what I suggested.

    For k < 1 the parabola y = f(x) opens downwards, so f is bounded from above (and can, therefore, have a finite maximum). For at least a range of k < 1, the equation f(x) = 0 will have two roots, and their mid-point will be the maximizing point of f(x). Just draw a picture of an upside-down parabola that cuts the x-axis in two points. You will then see what I am claiming.

    So, the maximum of ##ax^2 + bx + c## occurs at ##x^* = - b/(2a). ## In your case this becomes
    [tex] x^* = \frac{4k+3}{2(1-k)}[/tex] giving a maximum value of
    [tex] f_{\max} = f(x^*) = \frac{48k+1}{4(1-k)}.[/tex]
    (Note: this comes after a lot of simplification!) Since k < 1 the denominator is > 0, so the inequality we need is
    [tex] \frac{48k+1}{4(1-k)} \leq 12.5 \Longrightarrow 48k + 1 \leq 12.5 \times 4 (1-k),[/tex] or ##k \leq 1/2.##

    Note that the max, (48k+1)/[4(1-k)] decreases as k decreases from 1/2, and when k decreases down to k = -1/48 the two roots of the quadratic coallesce. For k < -1/48 the quadratic has no real roots at all, so f(x) < 0 for all x. The value ##f_{\max} = (48k+1)/[4(1-k)]## still applies in this case, but the calculus-free argument no longer applies and you need to use another way of seeing the result.
     
  11. Aug 13, 2013 #10

    Ray Vickson

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    Your error is going from
    [tex]\frac{2(k-1)(8k-4)-(4k+3)^2}{2(k-1)} \le 25[/tex]
    to
    [tex]2(8k^2-12k+4)-(16k^2+24k+9) \le 50(k-1)[/tex].
    This is false if k < 1, because in the fraction the denominator is < 0, so when you multiply through by it the inequality gets reversed: you should have
    [tex]2(8k^2-12k+4)-(16k^2+24k+9) \geq 50(k-1)[/tex].
     
  12. Aug 14, 2013 #11

    ehild

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    You made a mistake, the coefficient for x is 4k+3.
    Anyway, you made it very complicated. The quadratic expression (k-1)x2+(4k-3)x+4k-14.5 can not exceed or equal to zero. That is, the quadratic equation (k-1)x2+(4k-3)x+4k-14.5=0 has no roots. What is the condition for that?


    ehild
     
  13. Aug 15, 2013 #12
    Ray, thanks for taking the time to respond. Most helpful.

    ehild. You've spotted another set of typos, thanks. Yes, your suggestion offered a far more direct way to the result. The discriminant approach for roots that are not real didn't occur to me at the time of tackling this; but it's certainly something I'll consider in the future when doing these sorts of problems. Thanks for taking the time to respond.

    Problem solved.
     
  14. Aug 15, 2013 #13

    ehild

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    Look at the figure. A graph of a quadratic function can change sign only if the associated quadratic equation has two distinct roots. If the discriminant is less then zero, the function is either positive or negative. Remember, that is very useful.

    ehild
     

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