The range of the function f(x)=9^x-3^x+1 is

f(x) = 9^x - 3^x + 1

f(x) = (3^2x) - 3^x + 1

let 3^x = t

f(x) = t² - t + 1

f(x) = t² - t +1/4 - 1/4 + 1

f(x) = (t-1/2)² + 1-1/4

f(x) = (t-1/2)² + 3/4

square of any number cannot be less than 0 so minimum value of this function

is 3/4 when (t-1/2)² = 0

maximum value can be infinity

so its range is [3/4 , infinity)