How do you find the domain of #f(x) =1/sqrt(-x^2+5x-4)#?

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Answer 1 · 2017-08-25T20:07:39

The domain is #x in (1,4)#

For the square root and the division, the conditions are

#(-x^2+5x-4)>0# and #x!=1# and #x!=4#

Let #f(x)=(-x^2+5x-4)=-(x^2-5x+4)=-(x-1)(x-4)#

We can build the sign chart

#color(white)(aaaa)##x##color(white)(aaaaaaa)##-oo##color(white)(aaaaaa)##1##color(white)(aaaaaaaa)##4##color(white)(aaaaaa)##+oo#

#color(white)(aaaa)##-(x-1)##color(white)(aaaaa)##+##color(white)(aaa)##||##color(white)(aaa)##-##color(white)(aaaaaa)##-#

#color(white)(aaaa)##(x-4)##color(white)(aaaaaaa)##-##color(white)(aaa)####color(white)(aaaa)##-##color(white)(aa)##||##color(white)(aaa)##+#

#color(white)(aaaa)##f(x)##color(white)(aaaaaaaaaa)##-##color(white)(aa)##||##color(white)(aaa)##+##color(white)(aa)##||##color(white)(aaa)##-#

Therefore,

#f(x)>0#, when #x in (1,4)#

graph{1/sqrt(-x^2+5x-4) [-9.55, 18.93, -4.48, 9.76]}