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What is the range of (f(x)=\frac{4}{x^2+5}-1)?

Answer and explanation

Correct answer: ((-1,-\frac{1}{5}])

Direct answer: The range is \((-1,-\frac{1}{5}]\), so option A is correct. For every real x, x^2≥0, hence x^2+5≥5. Therefore 0<4/(x^2+5)≤4/5. The value 4/5 occurs at x=0. The fraction can approach 0 when |x| becomes very large, but it never equals 0, so its range is (0,4/5]. Subtracting 1 from every value shifts the interval left by 1, giving (-1, -1+4/5] = (-1,-1/5]. Option A has the correct open lower end and closed upper end. Option B wrongly includes −1, which is impossible. Option C is the range before the subtraction and misses the shift. Option D includes values below −1 that cannot occur. Memory cue: denominator is smallest at x=0, giving the maximum; infinity is approached but never reached.

Related tags

Real-Valued-FunctionsRangeRational-Shift

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