F(x) =√25-x² is increasing find x
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Step-by-step explanation:
Therefore,
#25-x^2>=0#
#(5+x)(5-x)>=0#
Let #f(x)=(5+x)(5-x)#
We can now build a sign chart
#color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaa)##-5##color(white)(aaaa)##5##color(white)(aaaa)##+oo#
#color(white)(aaaa)##5+x##color(white)(aaaaa)##-##color(white)(aaaa)##+##color(white)(aaaa)##+#
#color(white)(aaaa)##5-x##color(white)(aaaaa)##+##color(white)(aaaa)##+##color(white)(aaaa)##-#
#color(white)(aaaa)##f(x)##color(white)(aaaaaa)##-##color(white)(aaaa)##+##color(white)(aaaa)##-#
Therefore,
#f(x)>=0# when #x in [-5, 5]#
The domain of #f(x)# is #D_f(x)= [-5,5]#
When #x=+-5#, #sqrt(f(x))=0#
When #x=0#, #sqrt(f(x))=5#, this is the max. value
so,
The range is #R = [0,5]#
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