Question

Domain of the function :

Answer 1

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

Answer 2

5x−x 2>0

Answer 3

How?

Answer 4

what do you mean?

Answer 5

a.[1,4]
B.(1,4)
C.(0,5)
D.[0,5}

I mean can you show your work..

Answer 6

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

Answer 7

1,4 are the roots in the reals.

Answer 8

diya how come option c and d are same

Answer 9

they are not

Answer 10

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

Answer 11

Answer is A ..

Answer 12

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

Answer 13

oops,
you right. My bad

Answer 14

\[\sqrt{\log_{10}(\frac{5x-x^2}{4}) }=0\]
\[\frac{\sqrt{\ln (\frac14(5x-x^2))}}{\sqrt{\ln(10)}}=0\]
\[\sqrt{\ln(\frac14(5x-x^2))}=0\]
\[\ln(\frac14(5x-x^2))=0\]
\[\frac14(5x-x^2)=1\]
\[5x-x^2=4\]
\[x^2-5x=-4\]
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.

Answer 15

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

Answer 16

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

Answer 17

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

Answer 18

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))

Answer 19

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?

Answer 20

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

Answer 21

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

Answer 22

ok

Answer 23

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

Answer 24

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

Answer 25

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

Answer 26

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

Answer 27

Oh yeah ok

Answer 28

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 \[5x-x^2/4 >0\]

Answer 29

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

Answer 30

@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.

Answer 31

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

Answer 32

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

Answer 33

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 >.<

Answer 34

Ok thanks :D you're right

and one more thing Why are we taking equal to zero instead of greater than zero ?

Answer 35

If you include the values then the argument would be 0 and that is not possible. So it shall be (1,4) I believe

Answer 36

The argument of a logarithm is never 0

Answer 37

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

Answer 38

=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.

Answer 39

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

Answer 40

Ok Thank you so much :)

Answer 41

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