mathslover
  • mathslover
find the minimum value of \( 4-5x-4x^2\).
Mathematics
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mathslover
  • mathslover
find the minimum value of \( 4-5x-4x^2\).
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
For any parabola of the form \(ax^2+bx+c=0 \) the vertex will be at \( \large \left(-\frac b {2a} , -\frac {b^2-4ac}{4a}\right) \).Equation of line of symmetry will be at \(x=-\frac b {2a} \) Maximum and minimum values will be at \( \large \left(-\frac b {2a} , -\frac {b^2-4ac}{4a}\right) \) and \( \large \left(-\frac b {2a} , -\frac {b^2-4ac}{4a}\right) \) for \(a<0\) and \(a>0\) respectively.
mathslover
  • mathslover
@FoolForMath but that does not lead to the correct answer
anonymous
  • anonymous
i think there is no minimum value

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anonymous
  • anonymous
it's unbounded
anonymous
  • anonymous
For this problem there will be no minimum values.
mathslover
  • mathslover
can we do this like : 4-5x-4x^2 4-4(x^2+5x/4) 4-4(x^2+5x4 +25/64) + 25/16 ?
mathslover
  • mathslover
is this the correct method ?
anonymous
  • anonymous
|dw:1340428823754:dw|
anonymous
  • anonymous
just maximum value
mathslover
  • mathslover
4-4(x^2+5x4 +25/64) + 25/16 89/16-4((x+5/8)^2) hence it as min. value of 89/16 at x =-5/8 is it right ?
anonymous
  • anonymous
nope
mathslover
  • mathslover
why ?
anonymous
  • anonymous
THIS FUNCTION HAS NO MINIMUM VALUE
mathslover
  • mathslover
http://www.wolframalpha.com/input/?i=4-5x-4x%5E2&t=crmtb01 i am getting right according to this but it may be wrong please have a look
anonymous
  • anonymous
Lets try to plot this function graphically,
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mathslover
  • mathslover
k
anonymous
  • anonymous
ok try x = 10000
anonymous
  • anonymous
when do you think the minimum value will be achieved?
anonymous
  • anonymous
why?
mathslover
  • mathslover
can u tell m not soo good
anonymous
  • anonymous
every x you get i will pick one,that that my function will be less than yours
anonymous
  • anonymous
Yes, for every minimum you find I can find another x which gives a lesser value ...
mathslover
  • mathslover
thanks a lot every one

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