If f(x) = 1/x and g(x) = x/(x^2 + 1), for all x 0, what is the minim

Question

If  f(x)=\frac{1}{x}  and  g(x) = \frac{x}{(x^2+1)} , for all  x > 0 , what is the minimum value of  f(g(x)) ?

(A) 0

(B)  \frac{1}{2}

(C) 1

(D) \frac{3}{2}

(E) 2

BrushMyQuant · Accepted Answer

Nested Functions

f(x)=\frac{1}{x}  and  g(x) = \frac{x}{(x^2+1)}  and we need to find the minimum value of  f(g(x))

To find  f(g(x))  we need to substitute value of g(x) first. So lets do that

f(g(x))  =  f(\frac{x}{(x^2+1)})

To find value of  f(\frac{x}{(x^2+1)})  we need to replace x with  \frac{x}{(x^2+1)}  in f(x)

We get,  f(\frac{x}{(x^2+1)})  = 1 /  \frac{x}{(x^2+1)}  =  \frac{(x^2+1)}{x}  =  \frac{x^2}{x}  +  \frac{1}{x}  = x +  \frac{1}{x}

Now we can differentiate and get the answer but let's go with substitution.
x +  \frac{1}{x}  = 0 is not possible as x > 0

Now for any value of x > 1, like x = 1.1
x +  \frac{1}{x}  will be more than 2

Any value of x between 0 and 0.5 again, like x = 0.1
x +  \frac{1}{x}  will be more than 2

For values of x between 0.5 and 1, like x=0.6
x +  \frac{1}{x}  will be more than 2

So, Minimum value of x +  \frac{1}{x}  will be 2 when x = 1

So,  Answer will be E 
Hope it helps!

Watch the following video to learn the Basics of Functions and Custom Characters

https://www.youtube.com/watch?v=b7JoMX0gk8M

Adarsh_24 · Answer

I think we can use differentiation too.

(x^2 + 1)/x =x + 1/x = x+x^-1

differentiating and equating to 0

1+ -1 * x^-2 =0

x^-2 =1

x !=0

x^2=1
x=+1 since x>0

Ans=2

JT2916 · Answer

If x=+-1 based on differentiation method, How come x=2?

Bunuel · Answer

We are given that x > 0, so x = -1 is not valid.

Krunaal · Answer

Since it's given in the question that functions are valid for x > 0, we take the value x = 1 and substitute it in f(g(x)) =  \frac{x^2 + 1 }{ x}

So, x = 1, and the  f(g(x)) = 2  => which is the answer we were asked to find.

JT2916 · Answer

Thanks, missed it.

saynchalk · Answer

To minimise g(x), lowest value is 1,

g(x) = 1/2

f(g(x)) = 2

iamchinu97 · Answer

I have better method
our final equation is (x+1/x) and we want to minimize it.
now lets check options
if X+1/X = 0 this is not possible as X>0
if X+1/X=1/2 then we get quadratic eq 2x^2 - x + 2 = 0 now X to have real value >0 this quadratic must have b^2-4ac >=0

Now if u will check same for other options u will find b^2-4ac > 0 only for option E.

If f(x) = 1/x and g(x) = x/(x^2 + 1), for all x > 0, what is the minim

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