Applications of Differentiation

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Homework Help Overview

The discussion revolves around finding critical numbers of the function g(t) = sqrt(4-t) for t < 3. Participants are exploring the implications of differentiating a square root function and the conditions under which critical numbers can be identified.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the process of finding critical numbers, including taking derivatives and setting them to zero. There is uncertainty about handling the square root and the implications of squaring both sides of an equation. Questions arise regarding the definitions of terms like "critical number" and the behavior of the function within the specified interval.

Discussion Status

Some participants have attempted to manipulate the equation and express their confusion regarding the definitions and the steps taken. There is acknowledgment of the need for clarity on the problem statement and the nature of the function. Guidance has been offered regarding algebraic manipulations, but no consensus has been reached on the correct approach or interpretation.

Contextual Notes

Participants note that the function g(t) becomes complex for t > 4, which may influence the understanding of critical numbers in the context of the given interval. There is also mention of the need for a clear definition of "critical number" as it relates to maxima and minima.

BoogieL80
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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.
 
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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?
 
I'm sorry, what does RHS and LHS stand for?
 
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.
 
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...
 
BoogieL80 said:
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.
 
I think I forgot to factor in my process solving. I think I've figured it out. Thank you for your help :)
 
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...
 

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