Editing Concave function (section)

===Functions of a single variable===
# A [[differentiable function]] {{mvar|f}} is (strictly) concave on an [[interval (mathematics)|interval]] if and only if its [[derivative]] function {{mvar|f &prime;}} is (strictly) [[monotonically decreasing]] on that interval, that is, a concave function has a non-increasing (decreasing) [[slope]].<ref>{{Cite book| last=Rudin| first=Walter| title=Analysis| year=1976| pages= 101}}</ref><ref>{{Cite journal |last1=Gradshteyn|first1=I. S.| last2=Ryzhik|first2=I. M.| last3=Hays|first3=D. F.| date=1976-07-01| title=Table of Integrals, Series, and Products| journal=Journal of Lubrication Technology| volume=98|issue=3|pages=479| doi=10.1115/1.3452897|issn=0022-2305 |doi-access=free}}</ref>
# [[Point (geometry)|Points]] where concavity changes (between concave and [[convex function|convex]]) are [[inflection point]]s.<ref>{{Cite book|last=Hass, Joel |url=https://www.worldcat.org/oclc/965446428| title=Thomas' calculus| others=Heil, Christopher, 1960-, Weir, Maurice D.,, Thomas, George B. Jr. (George Brinton), 1914-2006.|date=13 March 2017| isbn=978-0-13-443898-6| edition=Fourteenth| location=[United States]| pages=203| oclc=965446428}}</ref>
# If {{mvar|f}} is twice-[[Differentiable function|differentiable]], then {{mvar|f}} is concave [[if and only if]] {{mvar|f &prime;&prime;}} is [[non-positive]] (or, informally, if the "[[acceleration]]" is non-positive). If {{mvar|f &prime;&prime;}} is [[negative numbers|negative]] then {{mvar|f}} is strictly concave, but the converse is not true, as shown by {{math|1=''f''(''x'') = &minus;''x''<sup>4</sup>}}.
# If {{mvar|f}} is concave and differentiable, then it is bounded above by its first-order [[Taylor approximation]]:<ref name=":0" /> <math display="block">f(y) \leq f(x) + f'(x)[y-x]</math>
# A [[Lebesgue measurable function]] on an interval {{math|'''C'''}} is concave [[if and only if]] it is midpoint concave, that is, for any {{mvar|x}} and {{mvar|y}} in {{math|'''C'''}} <math display="block"> f\left( \frac{x+y}2 \right) \ge \frac{f(x) + f(y)}2</math>
# If a function {{mvar|f}} is concave, and {{math|''f''(0) ≥ 0}}, then {{mvar|f}} is [[subadditivity|subadditive]] on <math>[0,\infty)</math>. Proof:
#* Since {{mvar|f}} is concave and {{math|1 ≥ t ≥ 0}}, letting {{math|1=''y'' = 0}} we have <math display="block">f(tx) = f(tx+(1-t)\cdot 0) \ge t f(x)+(1-t)f(0) \ge t f(x) .</math>
#* For <math>a,b\in[0,\infty)</math>: <math display="block">f(a) + f(b) = f \left((a+b) \frac{a}{a+b} \right) + f \left((a+b) \frac{b}{a+b} \right)
\ge \frac{a}{a+b} f(a+b) + \frac{b}{a+b} f(a+b) = f(a+b)</math>