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Find the domain and range of `f(x)=sqrt(4-16x^(2))`. |
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Answer» Correct Answer - Domain: [-1/2, 1/2], Range: [0, 2] `f(x)=y=sqrt(4-16x^(2))` We must have `4-16x^(2) ge 0 implies x^(2) le 1//4` `implies x in [-1//2,1//2]` Hence domain is `[-1//2,1//2]` Also, minimum value of function is 0 and maximum value 2 when `16x^(2)=4` Hence range is `[0,2]` |
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