Simulation · Maths · Class 11
Domain and range as shadows
From the lesson Domain and range of real functions in Relations and Functions. Change the values and watch what happens.
Domain and range as shadowsMaths · Class 11
The idea behind it
NCERT §2.4
- When a real function is given only by a formula, its domain is taken to be the largest set of real numbers for which the formula gives a real value.
- Two things shrink the domain: a denominator cannot be 0, and an even root such as √ needs a non-negative quantity under it.
- For f(x) = (x² + 2x + 1)/(x² − 8x + 12), the denominator is (x − 2)(x − 6), so the domain is R − {2, 6}.
- For f(x) = √(x − 3), we need x − 3 ≥ 0, so the domain is [3, ∞); the range is [0, ∞), because a square root is never negative and x = 3 + k² gives the value k for any k ≥ 0.
- For f(x) = √(25 − x²), we need x² ≤ 25, so the domain is [−5, 5]; the value is largest (5) at x = 0 and smallest (0) at x = ±5, so the range is [0, 5].
- The range is found by asking which outputs are actually reached. For f(x) = x² + 4 on R the range is [4, ∞); for f(x) = 5 − 2x with x > 1 the range is (−∞, 3), since x > 1 gives 2x > 2 and so 5 − 2x < 3.
- A restriction on x changes the range: x² on R has range [0, ∞), but x² on [−1, 2] has range [0, 4].