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The range of the function $f left parenthesis x right parenthesis equals square root of left parenthesis x minus 1 right parenthesis left parenthesis 3 minus x right parenthesis end root text is end text$

$[0, 1]$
$(- 1, 1)$
$(- 3, 3)$
$(- 3, 1)$

hint Hint:

Here we have to find the range of f(x)= $square root of open parentheses x minus 1 close parentheses open parentheses 3 minus x close parentheses end root$ . Just solve the function and find the solution then find the range for the function.

The correct answer is: $left square bracket 0 comma 1 right square bracket$

Here we have to find that the range of $square root of open parentheses x minus 1 close parentheses open parentheses 3 minus x close parentheses end root$
f (x) = $square root of open parentheses x minus 1 close parentheses open parentheses 3 minus x close parentheses end root$
= $square root of negative x 2 plus 4 x minus 3 end root$
= $square root of negative x 2 plus 4 x minus 4 plus 1 end root$
= $square root of 1 minus open parentheses x minus 2 close parentheses 2 end root$
The maximum value of f (x) is 1,
when (x – 2) = 0.
So, we can write Fmax= 1
Now,
It is minimum when,
(x – 2)2 = 1
(x – 2) = ± 1 [ since, x = √a², then x = ± a]
At, positive, x -2 = 1 => x = 3
And at negative, x -2 = -1 => x = 1,
Therefore, x = 3, x = 1
So, Minimum = $square root of 1$ – 1 = 0
Therefore, the Range = [0, 1].

In this question, we have to find the range of f(x)= $square root of open parentheses x minus 1 close parentheses open parentheses 3 minus x close parentheses end root$ . Here solve the function and find when function is at maximum and minimum.

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The range of the function

Here we have to find the range of f(x)= . Just solve the function and find the solution then find the range for the function.

The correct answer is:

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