Номер 25(а,в,д,е) пожалуйста

Решите задачу:

$g(x)=3x^2-1 \\\ g(-2)=3\cdot(-2)^2-1=3\cdot4-1=12-1=11 \\\ g(4)=3\cdot4^2-1=3\cdot16-1=48-1=47 \\\ g(5)=3\cdot5^2-1=3\cdot25-1=75-1=74$

$g(x)= \frac{6}{x+1} +3 \\\ g(-5)= \frac{6}{-5+1} +3= \frac{6}{-4} +3=-1.5+3=1.5 \\\ g(0)= \frac{6}{0+1} +3= \frac{6}{1} +3=6+3=9 \\\ g(0)= \frac{6}{ \frac{1}{5} +1} +3= \frac{6}{1.2} +3=5+3=8$

$g(x)= \sqrt{x^2-3x+1} \\\ g(- \frac{1}{3} )= \sqrt{(- \frac{1}{3} )^2-3\cdot(- \frac{1}{3} )+1} =\sqrt{ \frac{1}{9} +1+1}= \sqrt{ 2\frac{1}{9} } =\sqrt{ \frac{19}{9} } = \frac{ \sqrt{19} }{3} \\\ g(0 )= \sqrt{0^2-3\cdot0+1} =\sqrt{ 1}= 1 \\\ g(4 )= \sqrt{4^2-3\cdot4+1} =\sqrt{ 16-12+1}= \sqrt{5}$

$g(x)=3-\cos \frac{x}{2} \\\ g(0)=3-\cos \frac{0}{2} =3-\cos0=3-1=2 \\\ g( \frac{ \pi }{3} )=3-\cos \frac{ \pi /3}{2} =3-\cos \frac{ \pi }{6} =3- \frac{ \sqrt{3} }{2} = \frac{6- \sqrt{3} }{2} \\\ g( - \pi )=3-\cos \frac{- \pi }{2} =3-\cos \frac{ \pi }{2} =3-0=3$