The function
f(x) =
Rational function. A ratio of two polynomials, of degree 2 over degree 1. The numerator's degree is one higher, so expect an oblique asymptote.
Existence conditions and domain
Each part of the expression imposes a condition:
- x − 2 ≠ 0
Solving them together and keeping the overlap:
D = (−∞, 2) ∪ (2, +∞)
Symmetry and periodicity
f(−x) matches neither f(x) nor −f(x), so there is no symmetry about the y-axis or the origin.
Intercepts
f(0) = 0.5 — the graph crosses the y-axis at (0, 0.5).
Solving f(x) = 0 gives x = −1, 1
Sign of f(x)
Where the curve sits above or below the x-axis:
| (−∞, −1) | f(x) < 0 — below the axis |
| (−1, 1) | f(x) > 0 — above the axis |
| (1, 2) | f(x) < 0 — below the axis |
| (2, +∞) | f(x) > 0 — above the axis |
Limits at the edges of the domain
- limx→−∞ f(x) = −∞
- limx→2⁻ f(x) = −∞
- limx→2⁺ f(x) = +∞
- limx→+∞ f(x) = +∞
Asymptotes
- x = 2vertical
- lim x→2⁻ f(x) = −∞; lim x→2⁺ f(x) = +∞
- y = 1x + 2oblique
- m = lim x→+∞ f(x)/x = 1; q = lim x→+∞ [f(x) − mx] = 2
First derivative: where it rises and falls
f′(x) =
| (−∞, 0.2679) | f′ > 0 — increasing |
| (0.2679, 3.7321) | f′ < 0 — decreasing |
| (3.7321, +∞) | f′ > 0 — increasing |
- (0.2679, 0.5359) — local maximum
- (3.7321, 7.4641) — local minimum
Over the whole domain:
- No absolute maximum: the function is unbounded above.
- No absolute minimum: the function is unbounded below.
Second derivative: concavity
f″(x) =
| (−∞, 2) | f″ < 0 — concave down |
| (2, +∞) | f″ > 0 — concave up |
Concavity never changes, so there are no inflection points.
The graph
- f(x)
- asymptote
- maximum / minimum