• September 10th 2009, 03:25 PM
marie7
f(x) = 4x^3 - 8squarerootx for 0 equal/less x equal/less 10

I've got thus far
y' = 12x^2 - (4/squarerootx)
when y' = 0, x = 0.64439
y'' = 24x + 2x^-1.5
when x = 0.64439, y'' = 19.33, which is a local minimum
y-coordinate of x = 0.64439 is -5.35
Left boundary x = 0, y = 0
Right boundary x = 10, y = 3,974.70

and I know how to draw the graph

okay so what is the global minimum/maximum? What does that mean?
• September 10th 2009, 03:45 PM
pickslides
$f(x) = 4x^3 - \sqrt[8]{x} ~, ~0 \leq x \leq 10$
$f(x) = 4x^3 - 8\times \sqrt{x} ~ , ~0 \leq x \leq 10$