Applications of Differentiation

[BoogieL80]

Homework Statement

Find any critical numbers of the function

Homework Equations

g(t) = sqrt (4-t) , t < 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.

 

[berkeman]

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?

 

[BoogieL80]

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

 

[berkeman]

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.

 

[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...

 

[berkeman]

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...

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.

 

[BoogieL80]

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

 

[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...