Question

Find critical points
h(x)= root over(6-3x^2)

Answer 1

To find critical points, you have to differentiate h.
Then solve h'(x)=0.
The solutions are the critical points.

Answer 2

HI , i did and sort of stuck at the end -3x(6-3x^2)^-1/2=0, x=0, and (6-3x^2)^(-1/2)=0, so how do i do at the second part... thanks for the help.

Answer 3

can you retype the function? as written, this is what you have: h(x)= root over(6-3x^2)

Answer 4

OK, so you've got your derivative! Solving h'(x) = 0:
\[-\frac{ 3x }{ \sqrt{6-3x^2} }=0 \Leftrightarrow 3x=0\]
So x=0

Answer 5

Remember: If a/b=0 this means: a=0 (and b is not 0). So there is an easy solution here...

Answer 6

so, there is only one critical number (0)..

Answer 7

Yes, we can get no more from this one...

Answer 8

so what is critical point is sqrt 6 !

Answer 9

(0, sqrt6)

Answer 10

Here is the graph, it confirms what you found:

Answer 11

its webwork solution and it is not getting right..

Answer 12

Thanks!

Answer 13

what would be the critical point ?

Answer 14

(0, sqrt(6))

Answer 15

There is even a maximum in that point.

Answer 16

the answer (0, sqrt(6)) cameout wrong in the webwork..

Answer 17

But the real question now is: are you convinced that it is true?