Dear student f(x)=14-x2for x to be defined denominator cant be zeroso4-x2=0x2=4x=±2Domain=R--2,2y=f(x)=14-x24-x2=1y4-x2=1y2x2=4-1y2=4-y2y2x=±4-y2y2x is defined for all y except y=0y=f(x)=14-x2 which is always positiveRange = (0,∞)ii) f(x)=3-x-2f(x) is defined for all xDomain =Rmod is always positiveso maximum value of f(x)=3 , when x=2range=(-∞,3] Regards

Please solve Q5.

Dear student

f (x) = \frac{1}{\sqrt{4 - x^{2}}} f o r x t o b e d e f i n e d d e n o m i n a t o r c a n t b e z e r o s o 4 - x^{2} = 0 x^{2} = 4 x = \pm 2 D o m a i n = R - \{- 2, 2\} y = f (x) = \frac{1}{\sqrt{4 - x^{2}}} \sqrt{4 - x^{2}} = \frac{1}{y} 4 - x^{2} = \frac{1}{y^{2}} x^{2} = 4 - \frac{1}{y^{2}} = \frac{4 - y^{2}}{y^{2}} x = \pm \sqrt{\frac{4 - y^{2}}{y^{2}}} x i s d e f i n e d f o r a l l y e x c e p t y = 0 y = f (x) = \frac{1}{\sqrt{4 - x^{2}}} w h i c h i s a l w a y s p o s i t i v e R a n g e = (0, \infty) i i) f (x) = 3 - |x - 2| f (x) i s d e f i n e d f o r a l l x D o m a i n = R m o d i s a l w a y s p o s i t i v e s o m a x i m u m v a l u e o f f (x) = 3, w h e n x = 2 r a n g e = (- \infty, 3]

Regards

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