Question

Given the derivative: (8-3x) / (2sqrt (4-x) )

Find the local max and local min

Answer 1

I got x = 8/3

but then there are many possible numbers to make the denominator undefined..so are those still consider critical points?

Answer 2

i would not worry about that, since no doubt the orignal funciton would be undefined there as well. maybe only consider where the denominator is 0, because the square root of zero is zero, so no problem for the original function, but a critical point of the derivative because you cannot divide by 0

Answer 3

so should I just use 8/3 and find the local max/min?

Answer 4

so anything undefined in the derivative mean I don't need to calculate anything from there??

Answer 5

yes and you will see that at 8/3 the derivative goes from being postive to negativef, so you have a local max, because the function goes from increasing to decreasing

Answer 6

and here x=4 can make the denominator 0 though

Answer 7

yes and that will no doubt be the endpoint of the domain of your function

Answer 8

in fact, if you can integrate you will see that the original funciton is
\[f(x)=x\sqrt{4-x}\] although of course you could be off by a constant

Answer 9

ohh right, cuz anything beyond x=4 will make the function undefined, and since it is the end point, there cannot be a local max or min

Answer 10

zactly

Answer 11

ok thanks !!