SL 5.2—Increasing and decreasing functions

$f^{\prime }(x)=2x-4$.

Set $f^{\prime }(x)=0$: $2x-4=0\Rightarrow x=2$.

For $x<2$ (e.g. $x=0$): $f^{\prime }(0)=-4<0$. For $x>2$ (e.g. $x=3$): $f^{\prime }(3)=2>0$.

Since $f^{\prime }(x)>0$ for $x>2$, $f$ is increasing on $(2,\infty )$.

$(-\infty ,2)$ is where $f^{\prime }(x)<0$, so this is the decreasing interval, not increasing.

$(-\infty ,4)$ uses an incorrect critical point; the true critical point is $x=2$, not $x=4$.

$(4,\infty )$ also uses the wrong critical point $x=4$, likely from misreading a coefficient.

Only $(2,\infty )$ correctly reflects where $f^{\prime }(x)>0$.

$f^{\prime }(x)=0$ at $x=-2$ and $x=1$.

For $x=0$ in $(-2,1)$: $f^{\prime }(0)=(0-1)(0+2)=-2<0$.

For $x=-3$: $f^{\prime }(-3)=(-4)(-1)=4>0$. For $x=2$: $f^{\prime }(2)=(1)(4)=4>0$.

$f^{\prime }(x)<0$ only on $(-2,1)$, so $f$ is decreasing there.

$(-\infty ,-2)$ has $f^{\prime }(x)>0$, so $f$ is increasing there, not decreasing.

$(1,\infty )$ also has $f^{\prime }(x)>0$, so $f$ increases there.

This combines the two increasing intervals, the opposite of decreasing behaviour.

Only $(-2,1)$ satisfies $f^{\prime }(x)<0$.