Question

Let $f$ and $g$ be two differentiable functions defined from $R\to R^{+}$. If $f(x)$ has a local maximum at $x=c$ and $g(x)$ has a local minimum at $x=c$, then $h(x)=\frac{f(x)}{g(x)}$

• A. has a local maximum at $x=c$
• B. has a local minimum at $x=c$
• C. is monotonic at $x=c$
• D. has a point of inflection at $x=c$

Answer 1

Answer: (A)

Given, $h(x)=\frac{f(x)}{g(x)}$
$\Rightarrow h^{\prime }(x)=\frac{g(x)f^{\prime }(x)-f(x)g^{\prime }(x)}{g^2(x)}$
Clearly, $h^{\prime }(c)=0($ As, $f^{\prime }(c)=0$ and $g^{\prime }(c)=0)$
$\text{ So, }{h}^{\prime }\left(c^{-}\right)=\frac{g\left(c^{-}\right)f^{\prime }\left(c^{-}\right)-f\left(c^{-}\right)\cdot g^{\prime }\left(c^{-}\right)}{g^2\left(c^{-}\right)}\left[f^{\prime }\left(c^{-}\right)>0;g^{\prime }\left(c^{-}\right)<0;f\left(c^{-}\right)>0;g\left(c^{-}\right)>0\right]$
$\therefore h^{\prime }\left(c^{-}\right)>0$
$||{|}^{ly}{h}^{\prime }\left(c^{+}\right)<0$
So, $h(x)$ has a local maximum at $x=c$.