[tex]\it f(x)=2^x(x^2+1)-3\\ \\ f\Big(\dfrac{1}{2}\Big)=\sqrt2(\dfrac{1}{4}+1)-3=\dfrac{3\sqrt2}{4}-3=\dfrac{\sqrt{18}}{4}-3 < 0\\ \\ \\ f(1)=2\cdot2-3=1 > 0\\ \\ Deci,\ f(x)\ are\ un\ zerou\ dac\breve a\ x\in\Big(\dfrac{1}{2},\ 1\Big)\subset(0,\ 1)[/tex]
