Question

Find the interval in which the function $f\left(x\right)=10-6x-2x^2$ is strictly increasing or decreasing

Answer 1

$f\left(x\right)=10-6x-2x^2$
$\therefore f^{\prime }\left(x\right)=-6-4x$
Now, $f^{\prime }\left(x\right)=0\Rightarrow x=-\frac{3}{2}$
The point $x=-\frac{3}{2}$ divides the real line into two disjoint intervals i.e., $(-\infty ,-\frac{3}{2})$ and $(-\frac{3}{2},\infty )$.
In interval $(-\infty ,-\frac{3}{2})$ i.e., when $x<-\frac{3}{2},f^{\prime }\left(x\right)=-6-4x<0.$
$\therefore$ $f$ is strictly decreasing for $x<-\frac{3}{2}$ and in interval $(-\frac{3}{2},\infty ),f^{\prime }\left(x\right)>0$
Thus $f$ is strictly increasing in this interval.