Odpowiedź :
1.
a)
[tex]f(x)=-2x+3\\f\left(-1\cfrac{1}{2}\right)=-2\left(-\cfrac{3}{2}\right)+3=3+3=6\\f\left(\sqrt{2}+1\right)=-2\left(\sqrt{2}+1\right)+3=-2\sqrt{2}-2+3=1-2\sqrt{2}\\f\left(-4\right)=-2\left(-4\right)+3=8+3=11[/tex]
b)
[tex]f(x)=-x^2+4x-1\\f\left(-1\cfrac{1}{2}\right)=-\left(-1\cfrac{1}{2}\right)^2+4\left(-1\cfrac{1}{2}\right)-1=-\left(-\cfrac{3}{2}\right)^2+4\left(-\cfrac{3}{2}\right)-1\\=-\cfrac{9}{4}-\cfrac{12}{2}-1=-\cfrac{9}{4}-\cfrac{24}{4}-\cfrac{4}{4}=-\cfrac{37}{4}=-9\cfrac{1}{4}\\f\left(\sqrt{2}+1\right)=-\left(\sqrt{2}+1\right)^2+4\left(\sqrt{2}+1\right)-1=(2+2\sqrt{2}+1)\\+4\left(\sqrt{2}+1\right)-1=-2-2\sqrt{2}-1+4\sqrt{2}+1-1=-3+2\sqrt{2}\\f\left(-4\right)=-(-4)^2+4(-4)-1=-16-16-1=-33[/tex]
2.
a)
[tex]f(x)=-4\cfrac{2}{3}x-1\cfrac{1}{2}\\-4\cfrac{2}{3}x-1\cfrac{1}{2}=0,\ x\in\mathbb{R}\\-4\cfrac{2}{3}x=1\cfrac{1}{2}\\-\cfrac{28}{6}x=\cfrac{9}{6}\\28x=-9\\x=-\cfrac{9}{28}[/tex]
b)
[tex]f(x)=3x-2\cfrac{1}{6}\\3x-2\cfrac{1}{6}=0,\ x\in\mathbb{R}\\3x=2\cfrac{1}{6}\\3x=\cfrac{13}{6}\\x=\cfrac{13}{18}[/tex]
c)
[tex]f(x)=-2x^2+32\\-2x^2+32=0,\ x\in\mathbb{R}_-\\2x^2=32\\x^2=16\\x=-4\ \lor\ x=4\leftarrow \text{odpada}\\x=-4[/tex]