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

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### 2 Answers

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.**

You question is what is maximum value !

`f(x)=sqrt(1+xe^(-3x))`

differentiate f(x) with respect to x

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

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

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

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

`e^(-3x)!=0`

`1-3x=0`

`x=1/3`

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

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

`(d^2(f(x)))/(dx^2)`

`=1/(2(1+xe^(-3x)))(-3e^(-3x)sqrt(1+xe^(-3x))(2-3x)-(e^(-6x)(1-3x)^2)/(2sqrt(1+xe^(-3x))))`

`(d^2(f(x)))/(dx^2)=`

`(1/(2(1+xe^(-3x))))(sqrt(1+xe^(-3x))(-3e^(-3x)(2-3x))-((e^(-3x)(1-3x))^2)/(2sqrt(1+xe^(-3x))))`

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

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

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

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

`f(1/3)=sqrt(1+e^(-1)/3)`

=1.0595 (approx)

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