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The range of the function f(x)=9^x-3^x+1 is
f(x) = 9x - 3x + 1 f(x) = (32x) - 3x + 1 let 3x = 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)
f(x) = 9x - 3x + 1
f(x) = (32x) - 3x + 1
let 3x = 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)
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