IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

tarun_bits

Could anyone tell me points about how to find the range of a function.....not good AT it...good at domain not range..

Also tell range of following function

$y= \frac{1}{\sqrt{(x +3) (4- x)}}$

you can refer ML Khanna page 605 for this(2007 edition) ...i ma not clear with it

RATES ASSURED

@@@HAV A CHILLIN TIME@@@

srujana

$For\ a\ given\ function\ y=f(x),\\\ \\$

$Range\ is\ the\ set\ of\ all\ values\ and\ only\ those\ values\ which\ y=f(x)\ takes.\\\ \\\\$

$Given\ function,$

$y=\frac{1}{\sqrt{(x+3)(4-x)}}\\\ \\$

$On\ squaring,$

$y=\frac{1}{(x+3)(4-x)}\\\ \\$

$\Rightarrow x^2y-xy-12y+1=0\\\ \\$

$For\ the\ equation\ to\ have\ real\ values\Delta\ge0$

$\Rightarrow y^2\ge4(1-12y)$

$solve\ for\ y,\ you\ get\ all\ the\ possible\ values\ of\ y=f(x)$

$\ which\ is\ the\ range\ of\ f(x)$

savvej

denominator must be positive.

>: (x+3)(4-x)>0

>: (x+3)(x-4)<0

.: by wavy curve ,xbelongs to (-3,4).

budokai_tenkaichi_returns

nicely done

refer to goiit study material

savvej

@ srujana ,

see ur step the squaring one .

u forgot squaring y which will become y²

srujana

Oh...sry

However the procedure remains to be the same

tarun_bits

here ,,,,,,,,,,is my problem....didn`t undestand the step...

for equation to have real values...
.......plz do explain.........

srujana