Question

Find the critcal numbers of this following function:
y=x-sqrtx

Answer 1

Calc 1 right?

Answer 2

yes

Answer 3

Okay cool. Finding critical numbers is easy, it simply means points where the derivative is equal to 0. So take the derivative of that function and set it equal to 0, these x values will be your critical numbers.

Answer 4

I got this:
\[y'=1-1/2x ^{-1/2}\]
then I'm not sure how to get the critical numbers..

Answer 5

Technically the derivative would look like this.

\[y ' = 1 - \frac{ 1 }{ 2\sqrt{x}}\]

Then you set y' = 0 and solve for x

\[0 = 1 - \frac{ 1 }{ 2\sqrt{x}}\]

Answer 6

okay

Answer 7

then I got x=1/4 @Helloimjohn1234

Answer 8

Yup! Make sense?

Answer 9

But the answer x=1/4 and 0. Where did they get the zero from?

Answer 10

@Helloimjohn1234

Answer 11

I would argue that 0 is not a critical point, if you plug in 0 for x then y' = 1. Is it an online homework?

Answer 12

No, textbook work

Answer 13

Gotcha, just graphed the function to be sure and checked everything on a calculator. There's no way that 0 could be a critical point, here'a picture of the graph of the derivative

Answer 14

Okay! Thanks! :)