Math Geniuses, I need your help!?

f(x) = xe^(-ax)

Use the product and chain rules to find f '(x).

f '(x) = e^(-ax) - axe^(-ax)

Set f '(x) = 0 and solve for x to find the critical point.

0 = e^(-ax) - axe^(-ax)

0 = e^(-ax)(1 - ax)

e^(-ax) = 0 or 1 - ax = 0

e^(-ax) = 0 has no solutions so discard it

1 - ax = 0

1 = ax

x = 1/a

Compute f ''(1/a) to determine nature of the critical point.

f ''(x) = -ae^(-ax) - ae^(-ax) + a^2xe^(-ax)

f ''(x) = -2ae^(-ax) + a^2xe^(-ax)

f ''(1/a) = -2ae^(-a(1/a)) + a^2(1/a)e^(-a(1/a))

f ''(1/a) = -2a/e + a/e = -a/e

-a/e < 0 so we know a local maximum occurs at x = 1/a

Set f ''(x) = 0 to find points at which the second derivative is 0.

0 = -2ae^(-ax) + a^2xe^(-ax)

0 = e^(-ax)(-2a + a^2x)

e^(-ax) = 0 or -2a + a^2x = 0

e^(-ax) = 0 has no solution so discard it

-2a + a^2x = 0

a^2x = 2a

x = 2/a

Second derivative is 0 at (2/a, f(2/a)) = (2/a, xe^(-a(2/a))) = (2/a, xe^(-2))

Find the derivative of f.

f'(x)=x*(-a)*(e^(-ax))+e^(-ax)

=e^(-ax)*[-ax+1]

Equate to zero and note that e^(-ax) is never 0.

So 1-ax=0 or ax=1 or x=(1/a)

Thus the only stationary point for f is at x=(1/a).

To see if this stationary point is a locam maxima or minima or enither, we find the second derivative of f at that x.

Differentiating f'(x)=e^(-ax)(1-ax) with respect to x , we get

f''(x)=e^(-ax)*(-a)-a*e^(-ax)*(1-ax)

Setting x=1/a, f''(1/a)=e^(-1)*(-a)-0=(-a)e^(-1) which is <0 since a>0 and e^(-1)>0

Thus the stationary point is a local maximum.

Now, the second derivative is 0 when

f''(x)=e^(-ax)*(-a+1-ax)=0

which is when -a+1-ax=0

i.e., ax=1-a

or x=(1-a)/a