dear student f(x)=20xax is always positive for a>020xas x→-∞20x=20-∞=120∞=1∞=0so minimum value →0as x→∞20x=20∞=∞so range=(0,∞) regards

please answerthese

dear student

f (x) = 20^{x} a^{x} i s a l w a y s p o s i t i v e f o r a > 0 20^{x} a s x \to - \infty 20^{x} = 20^{- \infty} = \frac{1}{20^{\infty}} = \frac{1}{\infty} = 0 s o m i n i m u m v a l u e \to 0 a s x \to \infty 20^{x} = 20^{\infty} = \infty s o r a n g e = (0, \infty)

regards