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Learn how Kamakshi achieved a GMAT 675 with an impressive 96th %ile in Data Insights. Discover the unique methods and exam strategies that helped her excel in DI along with other sections for a balanced and high score.
26 Oct 2021, 06:15
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E
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Difficulty:
45%
(medium)
Question Stats:
67% (01:50) correct
33%
(01:59)
wrong
based on 1169
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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
(A) 0
(B) \(\frac{1}{2}\)
(C) 1
(D)\(\frac{3}{2}\)
(E) 2
26 Oct 2021, 09:32
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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
\(f(x)=\frac{1}{x}\) 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
General Discussion
25 Aug 2024, 03:29
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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
(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
