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ExamsJEE MainMaths

Determine the range of each function f(x) = x² - 8x + 7 for the following domains: (i) f: R -> R; (ii) f: [-1, 3] -> R; (iii) f: [-1, infinity) -> R; (iv) f: (-infinity, 3] -> R.

  1. (i) [-9, infinity); (ii) [-8, 16]; (iii) [-9, infinity); (iv) [-8, infinity)
  2. (i) [-9, infinity); (ii) [-8, 16]; (iii) [-8, infinity); (iv) [-9, infinity)
  3. (i) R; (ii) [0, 16]; (iii) [-9, infinity); (iv) [-9, infinity)
  4. (i) [-9, infinity); (ii) [-9, 16]; (iii) [-9, infinity); (iv) [-9, 16]

Correct answer: (i) [-9, infinity); (ii) [-8, 16]; (iii) [-9, infinity); (iv) [-8, infinity)

Solution

f(x) = (x-4)² - 9. (i) Over R the minimum is -9, range [-9, infinity). (ii) On [-1, 3] the vertex x=4 is outside, f decreasing on this interval: f(-1)=16, f(3)=-8, range [-8, 16]. (iii) On [-1, infinity) the vertex x=4 is included, min -9, so range [-9, infinity). (iv) On (-infinity, 3] the vertex is outside; f decreasing up to 3, f(3)=-8 is the minimum, range [-8, infinity).

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