Let f be a differentiable function on ] [α, β and ]. x ∈[α, β Show that, if
f ′(x) = 0 and ,0 f ′′(x) > then f must have a local maximum at x.
f′(x)=0,f′′(x)<0Since f′′is continuous on (α,β),f′′is negative on some interval (x−ε,x+ε)Taylor′s formula for ∣Δ∣<ε:f(x+Δ)=f(x)+f′(x)Δ+12f′′(ξ)Δ2=f(x)+12f′′(ξ)Δ2,ξ between x and x+Δf′′(t)<0,t∈(x−ε,x+ε)⇒f′′(ξ)<0⇒f(x+Δ)<f(x)⇒⇒x is a local maxf'\left( x \right) =0,f''\left( x \right) <0\\\\Since\,\,f'' is\,\,continuous\,\,on\,\,\left( \alpha ,\beta \right) , f'' is\,\,negative\,\,on\,\,some\,\,interval\,\,\left( x-\varepsilon ,x+\varepsilon \right) \\Taylor's\,\,formula\,\,for\,\,\left| \varDelta \right|<\varepsilon :\\f\left( x+\varDelta \right) =f\left( x \right) +f'\left( x \right) \varDelta +\frac{1}{2}f''\left( \xi \right) \varDelta ^2=f\left( x \right) +\frac{1}{2}f''\left( \xi \right) \varDelta ^2,\xi \,\,between\,\,x\,\,and\,\,x+\varDelta \\f''\left( t \right) <0,t\in \left( x-\varepsilon ,x+\varepsilon \right) \Rightarrow f''\left( \xi \right) <0\Rightarrow f\left( x+\varDelta \right) <f\left( x \right) \Rightarrow \\\Rightarrow x\,\,is\,\,a\,\,local\,\,\max \\f′(x)=0,f′′(x)<0Sincef′′iscontinuouson(α,β),f′′isnegativeonsomeinterval(x−ε,x+ε)Taylor′sformulafor∣Δ∣<ε:f(x+Δ)=f(x)+f′(x)Δ+21f′′(ξ)Δ2=f(x)+21f′′(ξ)Δ2,ξbetweenxandx+Δf′′(t)<0,t∈(x−ε,x+ε)⇒f′′(ξ)<0⇒f(x+Δ)<f(x)⇒⇒xisalocalmax
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