[tex]f(x)=x^3-2x^2+5x+1\\ f'(x)=3x^2-4x+5\\
3x^2-4x+5=0\\\Delta=(-4)^2-4\cdot3\cdot5=16-60=-44[/tex]
[tex]\Delta<0 \wedge a>0 \Rightarrow[/tex] the graph of the parabola is above the x-axis, so the derivative is always positive and therefore the initial function is increasing in its whole domain.
[tex]f(x)=0.5x^2-6\\
f'(x)=x[/tex]
The function is decreasing when its first derivative is negative. The first derivative of this function is negative for [tex]x<0[/tex] so for [tex]x\in(-\infty,0)[/tex] the function is decreasing.
[tex]f(x)=\dfrac{x+1}{x-1}\qquad(x\not=1)\\
f'(x)=\dfrac{x-1-(x+1)}{(x-1)^2}=-\dfrac{2}{(x-1)^2}[/tex]
The function is increasing when its first derivative is positive. The first derivative of this function is always negative therefore this function is never increasing.
