Determine the value of x where the maximum value of `f(x)=sqrt(1+(xe^(-3x)))`lies.  

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justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

The maximum value of a function f(x) is at the point x = a where f'(x) = 0 and f''(a) is negative.

For the function `f(x)=sqrt(1+(xe^(-3x)))` ,

f'(x) = `(1/2)*(1+(xe^(-3x)))^(-1/2)*(e^(-3x) + x*-3*e^(-3x))`

`(1/2)*(1+(xe^(-3x)))^(-1/2)*(e^(-3x) + x*-3*e^(-3x)) = 0`

=> `(e^(-3x) + x*-3*e^(-3x)) = 0`

=> `e^(-3x)(1 - 3x) = 0`

=> `x = 1/3`

f''(x) = `(sqrt(e^(3*x)+x)*((18*x-12)*e^(3*x)+9*x^2-6*x-1))/(4*e^(15*x/2)+8*x*e^(9*x/2)+4*x^2*e^(3*x/2))`

At x = 1/3, f''(x) `~~` -0.5208097 which is negative.

The maximum value of the the given function is at x = 1/3.

aruv's profile pic

aruv | High School Teacher | (Level 2) Valedictorian

Posted on

You question is what is maximum value !


differentiate f(x) with respect to x

`(d(f(x)))/(dx)=(1/2)((e^(-3x)-3xe^(-3x))) /sqrt(1+xe^(-3x))`


`` for maxima ,we equate `(d(f(x)))/(dx)=0`

`e^(-3x) (1-3x)=0--------(i)`




If this value of x=1/3 be point of maxima,we need to find second derivative of f(x).






`` because of (i) ,second term in bracket will zero.


which will always less than zero for x=1/3.

Thus x=1/3 will give maximum value,thus


=1.0595  (approx)

Thus maximum value of f(x)=1.0595 at point x=1/3.

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