Pulling fractional exponents out of an expression

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SUMMARY

The discussion focuses on finding critical numbers for the function F(t) = t^(3/4) - 2t^(1/4) by analyzing its derivative F'(t) = (1/4)t^(-3/4)(3t^(1/2) - 2). Participants emphasize the importance of correctly manipulating fractional exponents and suggest factoring t^(-3/4) from the derivative to simplify the expression. Additionally, the use of LaTeX for clarity in mathematical notation is recommended to avoid misinterpretation.

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  • Basic knowledge of factoring techniques in algebra
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  • Study the application of critical numbers in determining function behavior
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Cjosh
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Homework Statement


Find critical numbers of the function: F(x)=t^3/4 - 2t^1/4
Derivative I got: F'(x)=3/4 t^-1/4 - 1/2 t^-3/4

Homework Equations

The Attempt at a Solution


I have found the derivative and I understand I must pull out a t in order to find critical numbers, and run across this issue of manipulating fractional exponents too often. How do I go about this? Thankyou.
 
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Please post questions about calculus problems in the Calculus & Beyond section, not in the Precalculus section. I have moved your post.
Cjosh said:

Homework Statement


Find critical numbers of the function: F(x)=t^3/4 - 2t^1/4
Derivative I got: F'(x)=3/4 t^-1/4 - 1/2 t^-3/4
First off -- your two functions are functions of t, not x, so they should be written as F(t) and F'(t).

Cjosh said:

Homework Equations

The Attempt at a Solution


I have found the derivative and I understand I must pull out a t in order to find critical numbers, and run across this issue of manipulating fractional exponents too often. How do I go about this? Thankyou.
Rewrite the derivative so that it is a product rather than a sum. In this case, factor t^(-3/4) out of both terms.
 
Mark44 said:
Please post questions about calculus problems in the Calculus & Beyond section, not in the Precalculus section. I have moved your post.
First off -- your two functions are functions of t, not x, so they should be written as F(t) and F'(t).Rewrite the derivative so that it is a product rather than a sum. In this case, factor t^(-3/4) out of both terms.

So from this I get F'(t)= t^-3/4 (3/4t^1/2 - 1/2)
 
Cjosh said:
So from this I get F'(t)= t^-3/4 (3/4t^1/2 - 1/2)
Yes, that's correct, but even better is F'(t) = (1/4)t^(-3/4)[3t^(1/2) - 2)
Note that when you write exponents as inline text, t^-3/4 is usually interpreted as ##\frac{t^{-3}} 4##, using the usual rules of precedence.

This site supports the use of LaTeX (see https://www.physicsforums.com/help/latexhelp/ under the INFO menu, in Help/How-to articles).
Using LaTeX, the derivative looks like this:
$$F'(t) = \frac 1 4 t^{-3/4}(3t^{1/2} - 2)$$
 
Cjosh said:

Homework Statement


Find critical numbers of the function: F(x)=t^3/4 - 2t^1/4
Derivative I got: F'(x)=3/4 t^-1/4 - 1/2 t^-3/4

Homework Equations

The Attempt at a Solution


I have found the derivative and I understand I must pull out a t in order to find critical numbers, and run across this issue of manipulating fractional exponents too often. How do I go about this? Thankyou.

What you wrote means ##F = \frac{1}{4} t^3 - \frac{2}{4} t^1##. If you mean ##F = t^{3/4} - 2 t^{1/4}##, then you must use parentheses (or else use LaTeX, as I just did). With parentheses you would have F = t^(3/4) - 2 t^(1/4).
 

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