Let f be the function be defined by f(x)=5^((2x^2)
posted by Aparna on .
Let f be the function be defined by f(x)=5^((2x^2)1)^(1/2)
find the range of f without using a calculator

f(x)=5^((2x^2)1)^(1/2)
= 5^(x^2  1/2)
x^2 is always positive, so the smallest exponent that 5 could have is 1/2 when x = 0
for any other value of x, x^2  1/2 is greater than 1/2
we know 5^(1/2) = 1/√5 or √5/5
so the range is y ≥ √5/5