Question

Find the range of $f\left(x\right)=\sqrt{1-\sqrt{x^2-6x+9}}$.

Answer 1

f(x) = $\sqrt{1-\sqrt{x^2-6x+9}}$

f(x) = $\sqrt{1-\sqrt{(x-3{)}^2}}$

( in square root only positive quantities are defined so maximum value of

$\sqrt{(x-3{)}^2}$ we can take 1 when x=4,-1 and mininmum value of

$\sqrt{(x-3{)}^2}$ we can take is 0 when x=3)

so when $\sqrt{(x-3{)}^2}$ =1 than f(x) will be minimum and value =0

and when $\sqrt{(x-3{)}^2}$ =0 than f(x) will be maximum and value=1

so range of f(x) is [0, 1].