# Math Help - concavity

1. ## concavity

Im having trouble with finding the minimum value.

the function is (x)(e^5/x)

my derivative is (e^5/x)(-5x+1)

I set the derivative to 0 and get
x=-1/-5 (.2)

Plugging this back into the original equation gets me
14400979867.5...which is wrong. Any help would be appriciated

2. $f(x) = x * e^\frac{5}{x}$

Differentiating,

$\frac{df}{dx} = \frac{d}{dx} (x * e^\frac{5}{x}$

Applying the product rule,

$\frac{df}{dx} = e^\frac{5}{x} * \frac{d}{dx} (x) + x * \frac{d}{dx} (e^\frac{5}{x}$

By the chain rule,
$= \frac{df}{dx} = e^\frac{5}{x} + x (e^\frac{5}{x} * \frac{d}{dx}(\frac{5}{x})$

Which simplifies into

$f'(x) = e^\frac{5}{x} - \frac{5 * e^\frac{5}{x}}{x}$

Which can also be written as

$= e^\frac{5}{x} ( \frac{x -5}{x})$

So your derivative was wrong in the first place. Might be why your answer was coming up wrong

3. Originally Posted by Evan.Kimia
Im having trouble with finding the minimum value.

the function is (x)(e^5/x)

my derivative is (e^5/x)(-5x+1)

I set the derivative to 0 and get
x=-1/-5 (.2)

Plugging this back into the original equation gets me
14400979867.5...which is wrong. Any help would be appriciated
Shouldn't the derivative be (e^5/x)(-5x^(-1)+1)?

4. Hmm... so i tried 5 since x=0 does make y=0 but 0 isnt in the domain. I got 13.59~ for my min value but it didnt work

5. $x = 0$

Actually won't make f'(x) = 0.

$f'(x) = e^\frac{x}{5}(\frac{x - 5}{x})$

Plugging in x = 0,

$f'(0) = 1 (\frac{-5}{0})$

Which is clearly undefined. So, as you've said, x = 5 is your only critical point.

Now, taking the second derivative,

$f''(x) = \frac{25 * e^\frac{5}{x}}{x^3}$

It is readily verified that this function is only concave upwards, so there is only an absolute min.

6. $y = x e^{\frac{5}{x}}$

$y' = x e^{\frac{5}{x}} \cdot \left(-\frac{5}{x^2}\right) + e^{\frac{5}{x}}$

$y' = e^{\frac{5}{x}}\left(1 - \frac{5}{x}\right)$

y' is undefined at $x = 0$

for $x < 0$ , $y' > 0$ ... there are no extrema in this interval.

$y' = 0$ at $x = 5$

for $0 < x < 5$ , $y' < 0$

for $x > 5$ , $y' > 0$

a relative minimum exists at $x = 5$ , the value of which is $5e$.

The function has no absolute minimum. Consider what happens to the function as $x \to -\infty$.

7. Thanks, but when i plug x=5 back into my origional equation to get the min value i get 13.5914091423 which isnt correct

8. Did you copy the problem down correctly?

9. grrr... it wanted me to round it to 2 decimal places... thank you.