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The maximum value of 
is
is- a)(1/e)1/e
- b)(e)2/e
- c)(e)-1/e
- d)(e)1/e
Correct answer is option 'D'. Can you explain this answer?
Verified Answer
The maximum value ofisa)(1/e)1/eb)(e)2/ec)(e)-1/ed)(e)1/eCorrec...
For every real number (or) valued function f(x), the values of x which satisfies the equation f1(x)=0 are the point of it's local and global maxima or minima.
This occurs due to the fact that, at the point of maxima or minima, the curve of the function has a zero slope.
We have function f(x) = (1/x)x
We will be using the equation, y = (1/x)x
Taking in both sides we get
ln y = −xlnx
Differentiating both sides with respect to x.y.
dy/dx = −lnx−1
dy/dx =−y(lnx+1)
Equating dy/dx to 0, we get
−y(lnx+1)=0
Since y is an exponential function it can never be equal to zero, hence
lnx +1 = 0
lnx = −1
x = e(−1)
So, for the maximum value we put x = e^(−1)in f(x) to get the value of f(x) at the point.
f(e^−1) = e(1/e).
Hence the maximum value of the function is (e)1/e
This occurs due to the fact that, at the point of maxima or minima, the curve of the function has a zero slope.
We have function f(x) = (1/x)x
We will be using the equation, y = (1/x)x
Taking in both sides we get
ln y = −xlnx
Differentiating both sides with respect to x.y.
dy/dx = −lnx−1
dy/dx =−y(lnx+1)
Equating dy/dx to 0, we get
−y(lnx+1)=0
Since y is an exponential function it can never be equal to zero, hence
lnx +1 = 0
lnx = −1
x = e(−1)
So, for the maximum value we put x = e^(−1)in f(x) to get the value of f(x) at the point.
f(e^−1) = e(1/e).
Hence the maximum value of the function is (e)1/e
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The maximum value ofisa)(1/e)1/eb)(e)2/ec)(e)-1/ed)(e)1/eCorrec...
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Community Answer
The maximum value ofisa)(1/e)1/eb)(e)2/ec)(e)-1/ed)(e)1/eCorrec...
For every real number (or) valued function f(x), the values of x which satisfies the equation f1(x)=0 are the point of it's local and global maxima or minima.
This occurs due to the fact that, at the point of maxima or minima, the curve of the function has a zero slope.
We have function f(x) = (1/x)x
We will be using the equation, y = (1/x)x
Taking in both sides we get
ln y = −xlnx
Differentiating both sides with respect to x.y.
dy/dx = −lnx−1
dy/dx =−y(lnx+1)
Equating dy/dx to 0, we get
−y(lnx+1)=0
Since y is an exponential function it can never be equal to zero, hence
lnx +1 = 0
lnx = −1
x = e(−1)
So, for the maximum value we put x = e^(−1)in f(x) to get the value of f(x) at the point.
f(e^−1) = e(1/e).
Hence the maximum value of the function is (e)1/e
This occurs due to the fact that, at the point of maxima or minima, the curve of the function has a zero slope.
We have function f(x) = (1/x)x
We will be using the equation, y = (1/x)x
Taking in both sides we get
ln y = −xlnx
Differentiating both sides with respect to x.y.
dy/dx = −lnx−1
dy/dx =−y(lnx+1)
Equating dy/dx to 0, we get
−y(lnx+1)=0
Since y is an exponential function it can never be equal to zero, hence
lnx +1 = 0
lnx = −1
x = e(−1)
So, for the maximum value we put x = e^(−1)in f(x) to get the value of f(x) at the point.
f(e^−1) = e(1/e).
Hence the maximum value of the function is (e)1/e
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