Se considera functia f:\left(0,+\infty\right) definita prin f\left(x\right)=\frac{x^{4}}{4}-\ln x
a) Sa se calculeze f^{'}\left(x\right), x\in \left(0,+\infty\right)
b) Sa se determine punctele extreme ale functiei f.
c) Sa se demonstreze ca \ln\sqrt{x}\leq\frac{x^{2}-1}{4} pentru oricare x\in \left(0,+\infty\right).
Solutie

a) f^{'}\left(x\right)=\left(\frac{x^{4}}{4}-\ln x\right)^{'}=\frac{4x^{3}\cdot 4-x^{4}\cdot 0}{4^{2}}-\frac{1}{x}=\frac{16x^{3}}{16}-\frac{1}{x}=x^{3}-\frac{1}{x}=\frac{x\cdot x^{3}-1\cdot 1}{x}=\frac{x^{4}-1}{x}.
b) Stim ca f^{'}\left(x\right)=0, adica
\frac{x^{4}-1}{x}=0\Rightarrow \frac{\left(x^{2}-1\right)\left(x^{2}+1\right)}{x}=0, \forall x\in \left(0,+\infty\right)
si gasim
x^{2}+1=0\Rightarrow x^{2}=-1\Rightarrow x^{2}=i^{2}\Rightarrow x_{1,2}=\pm\sqrt{i^{2}}\Rightarrow x_{1,2}=\pm i\notin \left(0,+\infty\right) (deci nu convin).
x^{2}-1=0\Rightarrow x^{2}=1\Rightarrow x_{3,4}=\pm\sqrt{1}\Rightarrow x_{3,4}=\pm 1
Deci gasim x_{3}=1 si x_{4}=-1, observam ca x_{4} nu convine deoarece x se afla in intervalul \left(0, +\infty\right).
Acum trasam tabelul de variatie
monotonia unei functii
Calculam
f\left(1\right)=\frac{1^{2}}{4}-ln 1=\frac{1}{4}-0=\frac{1}{4}.
Deci f^{'}\left(x\right)\leq 0,\forall x\in\left(0, 1\right], adica functia este descrescatoare pe acest interval si f^{'}\left(x\right)\geq 0 \forall x\in \left[1,+\infty\right), adica functia este crescatoare pe acest interval. Si astfel gasim ca x=1 este punct de minim pentru functia f.
c) Din punctul b) stim ca x=1 este punct de minim global , deci \forall t>0, avem f\left(t\right)\geq f\left(1\right)\Rightarrow \frac{t^{4}}{4}-\ln t\geq \frac{1}{4}-\ln 1\Rightarrow \frac{t^{4}}{4}-\ln t\geq \frac{1}{4}\Rightarrow \frac{t^{4}}{4}-\frac{1}{4}\geq \ln t\Rightarrow \frac{t^{4}-1}{4}\geq \ln t.
Acum daca notam t=\sqrt{x} obtinem
\frac{\left(\sqrt{x}\right)^{4}-1}{4}\geq \ln \sqrt{x}\Rightarrow \frac{x^{2}-1}{4}\geq \ln \sqrt{x}.
Ceea ce trebuia demonstrat.
2) Se considera integrala I_{n}=\int^{2}_{1}x^{n}e^{x}dx, n\in N.
a) Sa se calculeze I_{0}
b) Sa se determine I_{1}
c) Sa se arate ca \left(n+1\right)I_{n}+I{n+1}=e\left(2^{n+1}e-1\right) pentru orice n\in N.
Solutie
a) I_{0}=\int^{2}_{1}x^{0}e^{x} dx=\int^{2}_{1}a\cdot e^{x}=\int^{2}_{1}e^{x} dx=e^{x}|^{2}_{1}=e^{2}-e^{1}=e^{2}-e=e\left(e-1\right)
b) I_{1}=\int^{2}_{1}x^{1}\cdot e^{x} dx=\int^{2}_{1}x\cdot e^{x}dx=\int^{2}_{1}x\cdot\left(e^{x}\right)^{'}dx=x\cdot e^{x}|^{2}_{1}-\int^{2}_{1}x^{'}\cdot e^{x}=x\cdot e^{x}|^{2}_{1}-\int^{2}_{1}1\cdot e^{x}dx=x\cdot e^{x}|^{2}_{1}-e^{x}|^{2}_{1}=2e^{2}-1\cdot e^{1}-e^{2}+e^{1}=e^{2}
Integrala de mai sus am rezolvat-o cu ajutorul integrarri prin parti.
c) I_{n+1}=\int^{2}_{1}x^{n+1}e^{x}dx=\int^{2}_{1}x^{n+1}e\left(e^{x}\right)^{'}dx=x^{n+1}e^{x}|^{2}_{1}-\int^{2}_{1}\left(x^{n+1}\right)^{'}e^{x}dx=2^{n+1}e^{2}-1^{n+1}e^{1}-\left(n+1\right)\int^{2}_{1}x^{n}e^{x}dx
Acum
I_{n}=\int^{2}_{1}x^{n}e^{x} dx
Acum obtinem
\left(n+1\right)I_{n}+I{n+1}=\left(n+1\right)\int^{2}_{1}x^{n}e^{x}dx+2^{n+1}e^{2}-1^{n+1}e^{1}-\left(n+1\right)\int^{2}_{1}x^{n}e^{x}dx=2^{n+1}e^{2}-1\cdot e=e\left(2^{n+1}e-1\right).

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