1. ## critical points

The following function on the domain t>0 has a local maximum at t=1/2. True of False?
$\displaystyle f(t) = \sqrt t .e^{ - t}$

When I differentiate I get the following, which I think is correct.
$\displaystyle f'(t) = e^{ - t} (\frac{1}{2}t^{\frac{{ - 1}}{2}} - t^{\frac{-1}{2}} )$

However I cannot see how t=1/2 is a critical point. I know critical point occur when 1st derivative is equal to zero however, I doesn't appear to me that t=1/2 makes f'(t) zero

2. Hello,
Originally Posted by Craka
The following function on the domain t>0 has a local maximum at t=1/2. True of False?
$\displaystyle f(t) = \sqrt t .e^{ - t}$

When I differentiate I get the following, which I think is correct.
$\displaystyle f'(t) = e^{ - t} (\frac{1}{2}t^{\frac{{ - 1}}{2}} - t^{\frac{-1}{2}} )$

However I cannot see how t=1/2 is a critical point. I know critical point occur when 1st derivative is equal to zero however, I doesn't appear to me that t=1/2 makes f'(t) zero
That's because it's not $\displaystyle f'(t) = e^{ - t} (\frac{1}{2}t^{\frac{{ - 1}}{2}} - t^{\frac{-1}{2}} )$

but rather $\displaystyle f'(t) = e^{-t} (\tfrac 12 t^{\frac{-1}{2}}-t^{\frac{{\color{red}+}1}{2}})$

$\displaystyle f'(t)=e^{-t} (\frac{1}{2 \sqrt{t}}-\sqrt{t})$

3. Thankyou for the reply, I can see where I made a mistake with differentiating. However I can still not see how there is a critical point at t=1/2. Maybe I missing something really elementary but could you point it out.

4. Originally Posted by Craka
Thankyou for the reply, I can see where I made a mistake with differentiating. However I can still not see how there is a critical point at t=1/2. Maybe I missing something really elementary but could you point it out.
$\displaystyle \frac{1}{2 \sqrt{t}}-\sqrt{t} = 0$

Multiply both sides by $\displaystyle 2 \sqrt{t}$:

$\displaystyle 1 - 2t = 0 \, ....$

5. $\displaystyle f'(t)=e^{-t} (\frac{1}{2 \sqrt{t}}-\sqrt{t})$

$\displaystyle 0 = (\frac{1}{2 \sqrt{t}}-\sqrt{t})$

$\displaystyle 0 = \frac{1 -2t}{2\sqrt{t}}$

$\displaystyle 0 = 1 -2t$

$\displaystyle -1 = -2t$

$\displaystyle \frac{1}{2} = t$

6. thankyou. Makes sense, only need to consider what was in brackets ie (1/2t^(-1/2)- t^1/2 ). As that will be the only time the function is able to be one as euler to any power is never zero right?