Diyadiya
  • Diyadiya
Domain of the function :
Mathematics
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chestercat
  • chestercat
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Diyadiya
  • Diyadiya
\[f(x)=\sqrt{\log_{10}(\frac{5x-x^2}{4}) }\]
anonymous
  • anonymous
5x−x 2>0
Diyadiya
  • Diyadiya
How?

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More answers

anonymous
  • anonymous
what do you mean?
Diyadiya
  • Diyadiya
a.[1,4] B.(1,4) C.(0,5) D.[0,5} I mean can you show your work..
anonymous
  • anonymous
5x>x2 5>x so i would say [0,5] since either logarithm or sqrt are not defind for negative values.
agreene
  • agreene
1,4 are the roots in the reals.
wasiqss
  • wasiqss
diya how come option c and d are same
anonymous
  • anonymous
they are not
Diyadiya
  • Diyadiya
its not same (1,5) and [1,5]
Diyadiya
  • Diyadiya
Answer is A ..
Mani_Jha
  • Mani_Jha
\[(\log_{10}(5x-x ^{2})/4 )\ge0\] \[(5x-x ^{2}/4)>0\] Intersection of the two solutions gives the answer.
anonymous
  • anonymous
oops, you right. My bad
agreene
  • agreene
\[\sqrt{\log_{10}(\frac{5x-x^2}{4}) }=0\] \[\frac{\sqrt{\ln (\frac14(5x-x^2))}}{\sqrt{\ln(10)}}=0\] \[\sqrt{\ln(\frac14(5x-x^2))}=0\] \[\ln(\frac14(5x-x^2))=0\] \[\frac14(5x-x^2)=1\] \[5x-x^2=4\] \[x^2-5x=-4\] solve that down u will find x=1 x=4 are the solutions, which is what I said the roots were these, and the answer is A.
agreene
  • agreene
^ that said, the function is really interesting in the complex plane.
anonymous
  • anonymous
I dont understand what are you doing with the 10, I thought it was the base of the log but it seems you are taking it as something different
Diyadiya
  • Diyadiya
How did you get \[\frac{\sqrt{\ln (\frac14(5x-x^2))}}{\sqrt{\ln(10)}}=0\] Sorry i forgot logs ..
agreene
  • agreene
i switched to ln cause it was easier to type, the bases really make no difference at all. but to get to the double radicals, you start by dividing by 1/sqrt(log(10))
anonymous
  • anonymous
If the base of the log is 10, then you should not write it even. So I dont understand the 10 there. Is the exercise good written?
Diyadiya
  • Diyadiya
Hm Yeah is given with base 10 ,But how does it make any difference ? base 10 or ln both are same ..@carolinekuhn
agreene
  • agreene
log base e is the natural log. What i did was this: log_10 (x)=ln(x)/ln(10)
Diyadiya
  • Diyadiya
ok
agreene
  • agreene
the first few steps were just to move to ln because logs not in base e are normally considered to be more complex than natural logs
Diyadiya
  • Diyadiya
\[\frac14(5x-x^2)=1\] how did you get this ? you took ln to the other side?
agreene
  • agreene
i took both sides to the e^() e^(ln(x)) = x and e^0 = 1
agreene
  • agreene
if you choose to stay with log_10 10^(log_10(x)) = x 10^0=1 it doesnt matter.
Diyadiya
  • Diyadiya
Oh yeah ok
anonymous
  • anonymous
You need that the whole expression is positive, because of the square root and then for the log to be positive the first condition is that argument of a log can only be positive, so you need to solve where \[5x-x^2/4 >0\]
Diyadiya
  • Diyadiya
Greater than zero or equal to zero ?which one should i use ?
agreene
  • agreene
@carolinekuhn that is extraneous to the derivation--if you happen to get results that fall outside that after derivation (ie: you check your answers and they are complex) then you must do that. In this case, the aforementioned derivation will allow you to determine all the real roots.
agreene
  • agreene
over the reals: http://www.wolframalpha.com/input/?i=plot+%5Csqrt%7B%5Clog_%7B10%7D%28%285x-x%5E2%29%2F%284%29%29+%7D%3D0
Diyadiya
  • Diyadiya
Can you tell me why we are taking [1,4] instead of (1,4) ?
agreene
  • agreene
I always get these confused, so i might be off on this.. but i believe [ and ] mean inclusive ( and ) mean non-inclusive that is to say: [1,4] means exactly 1 and 4 but (1,4) means close to but not including 1 or 4 [1,4) would be to include 1 but not 4 (but be very close to 4) unless I have this whole thing backwards again >.<
Diyadiya
  • Diyadiya
Ok thanks :D you're right and one more thing Why are we taking equal to zero instead of greater than zero ?
anonymous
  • anonymous
If you include the values then the argument would be 0 and that is not possible. So it shall be (1,4) I believe
anonymous
  • anonymous
The argument of a logarithm is never 0
Diyadiya
  • Diyadiya
But the answer given in my book is [1,4]
agreene
  • agreene
=0 will get you the roots (where the function crosses the x axis) because of the type of function, we knew it had to be a downward parabola (in the reals--due to sqrt and fraction) and we knew it would be contained therein (due to log) so from what we knew there--we knew to only look for the 2 roots for the domain.
anonymous
  • anonymous
oh yes I am cheking with 1 and 4 the result is 1 so you can include them. [1,4] is the answer
Diyadiya
  • Diyadiya
Ok Thank you so much :)
anonymous
  • anonymous
Diya, if you get this problem in a competitive you can use the options instead of the whole process ;)

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