# Applications of Differentiation

1. Mar 15, 2007

### BoogieL80

1. The problem statement, all variables and given/known data

Find any critical numbers of the function

2. Relevant equations
g(t) = sqrt (4-t) , t < 3

3. The attempt at a solution

I'm not really sure how to find a critical number when square roots are involved. I do know typically when finding the critical number you find the derivative of the equation and then set that equal to 0. However I ended up with the derivative dx = 4 sqrt (4-t) / 4-t. I tried setting that to 0 but got no where.

2. Mar 15, 2007

### Staff: Mentor

Hint: If SQRT(X) = 0, then X = 0 too, right? What can you do to the RHS to get rid of the square root? What do you get on the LHS when you do the same thing to it?

3. Mar 15, 2007

### BoogieL80

I'm sorry, what does RHS and LHS stand for?

4. Mar 15, 2007

### Staff: Mentor

No, that's my bad. I normally would have defined them with their first use if I wasn't sure you knew already.

LHS = left hand side (of the equation)
RHS = right hand side.

When working with any equation, you must do the same thing to the LHS as the RHS as you perform algebraic manipulations.

5. Mar 15, 2007

### BoogieL80

Okay, I tried squaring both sides and I got 16/(4-t) = 0. Is that t suppose to still be in the problem? I'm guessing that I have to eventually solve for t since the answer is t = 8/3. I'm assuming maybe I've done this problem incorrectly....

6. Mar 15, 2007

### Staff: Mentor

More likely, I'm not understanding the problem statement. Sorry, exactly how is the term "critical number" defined? I'd started off thinking maxima and minima, but that doesn't appear to be the case. BTW, they should be defining what interval of t they intend for the problem statement...

$$g(t) = \sqrt {4-t}$$

g(t) goes complex for t>4, I believe.

7. Mar 15, 2007

### BoogieL80

I think I forgot to factor in my process solving. I think I've figured it out. Thank you for your help :)

8. Mar 16, 2007

### end3r7

Hmm... you sure that's the right equation?.
The function $$g(t) = \sqrt {4-t}$$ is monotone decreasing on I:-infinity<t<3 and its derivative is defined and nonzero everywhere on the interval...