Question

Find range of (√(x-3)(5-x))?

Answer 1

we have to find range of $f(x)=\sqrt{(x-3)(5-x)}$

here function is defined only when
(x- 3)(5 - x) ≥ 0
or, 3 ≤ x ≤ 5
hence, domain of function is [3, 5]

now, differentiate function with respect to x,

f'(x) = 1/{√(x - 3)(5 - x)} × {(5 - x) - (x - 3)}
= (5 - x - x + 3)/√{(x - 3)(5 - x)}
= (8 - 2x)/√{(x - 3)(5 - x)}
at f'(x) = 0, 8 - 2x = 0 => x = 4

so, find value at x = 3, 4, 5

f(3) = √{(3 - 3)(5 - 3)} = 0
f(4) = √{(4 - 3)(5 - 4)} = 1
f(5) = √{(5 - 3)(5 - 5)} = 0

greatest value of f(x) = 1 and smallest value of f(x) = 0,

so, range is [0, 1]

Answer 2

Step-by-step explanation:

we have to find range of f(x)=\sqrt{(x-3)(5-x)}f(x)=

(x−3)(5−x)

here function is defined only when

(x- 3)(5 - x) ≥ 0

or, 3 ≤ x ≤ 5

hence, domain of function is [3, 5]

now, differentiate function with respect to x,

f'(x) = 1/{√(x - 3)(5 - x)} × {(5 - x) - (x - 3)}

= (5 - x - x + 3)/√{(x - 3)(5 - x)}

= (8 - 2x)/√{(x - 3)(5 - x)}

at f'(x) = 0, 8 - 2x = 0 => x = 4

so, find value at x = 3, 4, 5

f(3) = √{(3 - 3)(5 - 3)} = 0

f(4) = √{(4 - 3)(5 - 4)} = 1

f(5) = √{(5 - 3)(5 - 5)} = 0

greatest value of f(x) = 1 and smallest value of f(x) = 0,

so, range is [0, 1]