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Joined: 20 Feb 2017
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If f(x) = 1/x and g(x) = x/(x^2 + 1), for all x > 0, what is the minim
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06 Sep 2021, 09:46
06 Sep 2021, 09:46
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Question Stats:
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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
Re: If f(x) = 1/x and g(x) = x/(x^2 + 1), for all x > 0, what is the minim
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07 Sep 2021, 11:29
07 Sep 2021, 11:29
2
\(f(g(x)) = (x^2 + 1)/x\)
=\(f(g(x)) = x + 1/x\)
approach 1:
to find the minimum, take the first derivative and equate to 0
\(f'(g(x)) = 1 - 1/x^2 = 0\)
\(=> x = 1\)
so at \( x = 1 \) the value of \(f(g(x)) = 2\)
approach 2:
just based on the equation \(f(g(x)) = x + 1/x\)
x is a line with slope 1 and 1/x is curve and when added, we can note that it's minimum at x = 1 and raises immediately.
=\(f(g(x)) = x + 1/x\)
approach 1:
to find the minimum, take the first derivative and equate to 0
\(f'(g(x)) = 1 - 1/x^2 = 0\)
\(=> x = 1\)
so at \( x = 1 \) the value of \(f(g(x)) = 2\)
approach 2:
just based on the equation \(f(g(x)) = x + 1/x\)
x is a line with slope 1 and 1/x is curve and when added, we can note that it's minimum at x = 1 and raises immediately.
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Re: If f(x) = 1/x and g(x) = x/(x^2 + 1), for all x > 0, what is the minim
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10 Dec 2021, 10:30
10 Dec 2021, 10:30
1
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
