Question

Let $f$ and $g$ be two functions defined on an interval I such that $f\left(x\right)$ $\ge$ $0$ and $g\left(x\right)$ $\le$ $0$ for all $x$ $ϵ$ $I$ and $f$ is strictly decreasing on $I$ while $g$ is strictly increasing on $I$ then

• A. The product function $fg$ is strictly increasing on $I$
• B. The product function $fg$ is strictly decreasing on $I$
• C. $Fog\left(x\right)$ is monotonically increasing on $I$
• D. $Fog\left(x\right)$ is monotonically decreasing on $I$

Answer 1

Answer: (A), (D)

The product function $fg$ is strictly increasing on $I$
$Fog\left(x\right)$ is monotonically decreasing on $I$

Consider $f\left(x\right)=1-x$ and $g\left(x\right)=-x^2$
And let the interval be $Iϵ(-\infty ,0]$
Hence in the given interval $g\left(x\right)\le 0$ and $f\left(x\right)\ge 0$
Hence, $f^{\prime }\left(x\right)=-1$
$f^{\prime }\left(x\right)<0$, Hence it is decreasing.
$g^{\prime }\left(x\right)=-2x$
Now
$g^{\prime }\left(x\right)>0$ implies $2x<0$
$\therefore xϵ(-\infty ,0)$.
Thus in the given interval $g\left(x\right)$ is increasing while $f\left(x\right)$ is decreasing.
Now, $f\left(x\right).g\left(x\right)$ $=h\left(x\right)$
$=-x^2(1-x)$
$=x^3-x^2$.
$h^{\prime }\left(x\right)$ $=3x^2-2x>0$
$x(3x-2)>0$
$x<0$ or $x>\frac{2}{3}$
Now the given interval is $(-\infty ,0]$>
Hence $h\left(x\right)$ is increasing in that interval.
Thus $f\left(x\right).g\left(x\right)$ is increasing in that interval.
$f\left(g\right(x\left)\right)$ $=m\left(x\right)$
$=1-(-x^2)$
$=1+x^2$
Now, $m^{\prime }\left(x\right)=2x$
$m^{\prime }\left(x\right)<0$ for all $x<0$.
Hence, $f\left(g\right(x\left)\right)$ is decreasing in the given interval.