Domain of the function :

- Diyadiya

Domain of the function :

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- Diyadiya

\[f(x)=\sqrt{\log_{10}(\frac{5x-x^2}{4}) }\]

- anonymous

5x−x 2>0

- Diyadiya

How?

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## More answers

- anonymous

what do you mean?

- Diyadiya

a.[1,4]
B.(1,4)
C.(0,5)
D.[0,5}
I mean can you show your work..

- anonymous

5x>x2
5>x
so i would say
[0,5]
since either logarithm or sqrt are not defind for negative values.

- agreene

1,4 are the roots in the reals.

- wasiqss

diya how come option c and d are same

- anonymous

they are not

- Diyadiya

its not same (1,5) and [1,5]

- Diyadiya

Answer is A ..

- Mani_Jha

\[(\log_{10}(5x-x ^{2})/4 )\ge0\]
\[(5x-x ^{2}/4)>0\]
Intersection of the two solutions gives the answer.

- anonymous

oops,
you right. My bad

- 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

^ that said, the function is really interesting in the complex plane.

- 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

How did you get
\[\frac{\sqrt{\ln (\frac14(5x-x^2))}}{\sqrt{\ln(10)}}=0\]
Sorry i forgot logs ..

- 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

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

Hm Yeah is given with base 10 ,But how does it make any difference ? base 10 or ln both are same ..@carolinekuhn

- agreene

log base e is the natural log.
What i did was this:
log_10 (x)=ln(x)/ln(10)

- Diyadiya

ok

- 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

\[\frac14(5x-x^2)=1\]
how did you get this ? you took ln to the other side?

- agreene

i took both sides to the e^()
e^(ln(x)) = x
and
e^0 = 1

- agreene

if you choose to stay with log_10
10^(log_10(x)) = x
10^0=1
it doesnt matter.

- Diyadiya

Oh yeah ok

- 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

Greater than zero or equal to zero ?which one should i use ?

- 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

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

Can you tell me why we are taking [1,4] instead of (1,4) ?

- 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

Ok thanks :D you're right
and one more thing Why are we taking equal to zero instead of greater than zero ?

- 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

The argument of a logarithm is never 0

- Diyadiya

But the answer given in my book is [1,4]

- 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

oh yes I am cheking with 1 and 4 the result is 1 so you can include them. [1,4] is the answer

- Diyadiya

Ok Thank you so much :)

- anonymous

Diya, if you get this problem in a competitive you can use the options instead of the whole process ;)

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