# Thread: Find the absolute minimum

1. ## Find the absolute minimum

I need to find the absolute minimum value of $f(x)= 8x+ (1/16x^2)$ on the interval (0, infinity).

I took the first derivative, $f'(x)= 8-(1/8x^3)$ and set it equal to zero.

2. Originally Posted by rust1477
I need to find the absolute minimum value of $f(x)= 8x+ (1/16x^2)$ on the interval (0, infinity).

I took the first derivative, $f'(x)= 8-(1/8x^3)$ and set it equal to zero.

May I ask how you got $x=4?$ Check your work: $8-\frac1{8\cdot4^3}=8-\frac1{512}=\frac{4095}{512}\neq0.$

Try again.

3. Originally Posted by rust1477
I need to find the absolute minimum value of $f(x)= 8x+ (1/16x^2)$ on the interval (0, infinity).

I took the first derivative, $f'(x)= 8-(1/8x^3)$ and set it equal to zero.

Go back and re-work - carefully - your solution to f'(x) = 0.

4. ## Reworked f'(x)

I reworked the problem and got $(-1/8x^3)=-8$ and x=2 but I still don't think that 2 is the absolute minimum value. How should I check it?

5. You're wrong again...

$8-\frac{1}{8x^{3}}=0$

$\frac{1}{8x^{3}}=8$

$64x^{3}=1$

$x^{3}=\frac{1}{64}$

$x=\frac{1}{4}$

6. f'(2) does not equal 0
check again