SOLUTION: Find the domain and range of the following function in interval form.
f(x)= {{{ sqrt( x^2+3x-4 ) }}}* {{{ sqrt( 2-x-x^2 ) }}}
Algebra.Com
Question 1003122: Find the domain and range of the following function in interval form.
f(x)= *
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
without graphing these equations, you wouldn't have been able to see what is going on.
first i graphed sqrt(x^2 + 3x - 4)
that graph is shown below:
then i graphed sqrt(2 - x - x^2)
that graph is shwon below:
then i graphed both together on the same graph.
that graph is shown below:
then i graphed sqrt(x^2 + 3x - 4) * (2 - x - x^2)
that graph is shown below:
what happens is that the last graph shows nothing.
the reason for this is that the only valid interval for the graph of (2 - x - x^2) is within the invalid interval of (x^2 + 3x - 4) so it disappears.
the exclusion zone for the first expression is -4 < x < 1.
the exclusion zone for the second expression is x < -2 and x > 1.
put those 2 exclusion zones together and you have the entire domain of x being excluded which results in a null graph.
it took a little bit of analysis to figure this out algebraically.
first the expressions needed to be factored.
then the factors were analyzed to determine when the value under the square root sign became negative.
in order for sqrt((x+4)*(x-1)) to be negative, (x+4) had to be positive while (x-1) was negative and (x+4) had to be negative when (x-1) was positive.
in order for sqrt((2+x)*(1-x)) to be negative, (2+x) has to be positive while (1-x) was negative and (2+x) had to be negative when (1-x) was positive.
the analysis confirmed what was shown on the graph.
sqrt((x+4)*(x-1)) was negative when -4 < x < 1
sqrt((2+x)*(1-x)) was negative when x < -2 and x > 1.
sqrt((x+4)*(x-1)) was positive when x <= -4 and x >= 1
sqrt((2+x)*(1-x)) was positive when -2 <= x < 1
when you put them together on the same grpah, their exclusion zones together comprised all values of x so there were not values of x that could be graphed.
the domain was therefore null.
that's what i believe is happening.
the graphs confirmed it.
the analysis was fairly length but and a little bit complex, but it confirmed what the graphs showed.
i didn't show the analysis because it would have complicate the results of the analysis unnecessarily and i didn't have enough time to finish it.
if you need a better explanation, then write and i'll take the time tomorrow or the next day to work it out for you.
you should, however, be able to work it out for yourself, now that you know what's happening and how you need to proceed.
the graphing software i used is at link to calculator used
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