Question

find the domain and range of function f(x) = √(21-4x-x2)plzz answer

Answer 1

Answer:

Step-by-step explanation:

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Answer 2

Answer:

Domain of f(x) is [-7,3] , Range of f(x) is [-5,5]

Step-by-step explanation:

Given

$f(x) = \sqrt{21 - 4x - {x}^{2} }$

for finding the domain,the term under the square root must be greater than or equal to zero.

Hence,equation the above term we get the following inequality.

$21 - 4x - {x}^{2} ≥0 = > {x}^{2} + 4x - 21≤0$

by factorising the equation,we get the following inequality

$(x + 7)(x - 3)≤0 \\ = = > - 7≤x≤3$

Therefore,domain of f(x) is [-7,3]

to find range of f(x) we require f`(x)

$f`(x) = \frac{1}{2 \sqrt{21 - 4x - {x}^{2} } } ( - 4 - 2x)$

equate this to zero

$= = > -2 - x = 0 = = > x = - 2$

substitute x=-2 in given function

$f( - 2) = \sqrt{21 - 4( - 2) - {( - 2)}^{2} } = \sqrt{25 } = ±5$

hence minimum and maximum values if f(x) are -5,5

Therefore Range of f(x) is [-5,5]

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