recategorized by
459 views
1 votes
1 votes

Suppose that the function $h(x)$ is defined as $h(x)=g(f(x))$ where $g(x)$ is monotone increasing, $f(x)$ is concave, and $g’’(x)$ and $f’’(x)$ exist for all $x$. Then $h(x)$ is

  1. always concave
  2. always convex
  3. not necessarily concave
  4. None of these
recategorized by

1 Answer

0 votes
0 votes

The correct answer is option C.

Consider f(x) is a function that is twice continuously differentiable on an interval I. Then the function f(x) is 

  1. convex if $f''(x)>0 ,$ for all x in I.
  2. concave if $f''(x)<0 ,$for all x in I.

let take ,

$f(x)=-x^{2}$  

This graph is a concave graph as 

$f''(x)=-2<0$

Figure,

now ,

Let’s take $g(x)=e^{x}$ ,It is monotone increasing function.

Figure,

Now if we see $h(x)=g(f(x))=e^{-x^{2}}$

$h''(x)=2e^{-x^{2}}\left [ 2x^{2}-1 \right ]$.

It is concave when $-\frac{1}{\sqrt{2}}<x<\frac{1}{\sqrt{2}}$ range

it is convex when $x>\frac{1}{\sqrt{2}}$ and $x<-\frac{1}{\sqrt{2}}$

Graph ,

As concaveness or convexnees we can find out by second differentiation . 

As it is clear that it will never always concave or convex so option a and b is false.

 

 

Related questions

3 votes
3 votes
1 answer
1
Arjun asked Sep 23, 2019
497 views
It is given that $e^a+e^b=10$ where $a$ and $b$ are real. Then the maximum value of $(e^a+e^b+e^{a+b}+1)$ is$36$$\infty$$25$$21$
1 votes
1 votes
1 answer
2
Arjun asked Sep 23, 2019
548 views
The function $f(x) = x^{1/x}, \: x \neq 0$ hasa minimum at $x=e$;a maximum at $x=e$;neither a maximum nor a minimum at $x=e$;None of the above
0 votes
0 votes
0 answers
3
Arjun asked Sep 23, 2019
388 views
Let $f(x)=\sin x^2, \: x \in \mathbb{R}$. Then$f$ has no local minima$f$ has no local maxima$f$ has local minima at $x=0$ and $x=\pm\sqrt{(k+\frac{1}{2} ) \pi}$ for odd i...
0 votes
0 votes
1 answer
4
Arjun asked Sep 23, 2019
415 views
The function $f(x)=\sin x(1+ \cos x)$ which is defined for all real values of $x$has a maximum at $x= \pi /3$has a maximum at $x= \pi$has a minimum at $x= \pi /3$has neit...