Calculus Problem with Exponents

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Jason76

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\(\displaystyle f(t) = t^{3/4} - 3t^{1/4} \)

\(\displaystyle f'(t) = \dfrac{3}{4}t^{-1/4} - \dfrac{1}{4}3t^{3/4}\)

\(\displaystyle f'(t) = \dfrac{3}{4}t^{-1/4} - \dfrac{3}{4}t^{3/4}\)

\(\displaystyle f'(t) = \dfrac{3}{4}t^{-1/4} - \dfrac{3}{4}t^{3/4} = 0\)

\(\displaystyle f'(t) = \dfrac{3}{4}t^{-1/4}= \dfrac{3}{4} 3t^{3/4}\)

\(\displaystyle f'(t) = t^{-1/4}= t^{3/4}\) :confused: What do to do here. to get a single value \(\displaystyle t = ?\)
 
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Jason76 said:
\(\displaystyle f(t) = t^{3/4} - 3t^{1/4} \)

\(\displaystyle f'(t) = \dfrac{3}{4}t^{-1/4} - \dfrac{1}{4}3t^{3/4}\)....................Incorrect

\(\displaystyle f'(t) = \dfrac{3}{4}t^{-1/4} - \dfrac{3}{4}t^{3/4}\)

\(\displaystyle f'(t) = \dfrac{3}{4}t^{-1/4} - \dfrac{3}{4}t^{3/4} = 0\)

\(\displaystyle f'(t) = \dfrac{3}{4}t^{-1/4}= \dfrac{3}{4} 3t^{3/4}\)

\(\displaystyle f'(t) = t^{-1/4}= t^{3/4}\)

:confused: What do to do here. to get a single value \(\displaystyle t = ?\)
Click to expand...

t-1/4 = 1/(t1/4)
 
\(\displaystyle f(t) = t^{3/4} - 3t^{1/4} \)

\(\displaystyle f'(t) = \dfrac{3}{4}t^{-1/4} - \dfrac{1}{4}3t^{3/4}\)

\(\displaystyle f'(t) = \dfrac{3}{4}t^{-1/4} - \dfrac{3}{4}t^{3/4}\)

\(\displaystyle \dfrac{3}{4}t^{-1/4} - \dfrac{3}{4}t^{3/4} = 0\)

\(\displaystyle \dfrac{3}{4}t^{-1/4}= \dfrac{3}{4} 3t^{3/4}\)

\(\displaystyle t^{-1/4}= t^{3/4}\)

\(\displaystyle 1 = \dfrac{t^{3/4}}{t^{4}}\)
 
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Jason76 said:
\(\displaystyle f(t) = t^{3/4} - 3t^{1/4} \)

\(\displaystyle f'(t) = \dfrac{3}{4}t^{-1/4} - \dfrac{1}{4}3t^{3/4}\) ......................... Incorrect

\(\displaystyle f'(t) = \dfrac{3}{4}t^{-1/4} - \dfrac{3}{4}t^{3/4}\)

\(\displaystyle f'(t) = \dfrac{3}{4}t^{-1/4} - \dfrac{3}{4}t^{3/4} = 0\)

\(\displaystyle f'(t) = \dfrac{3}{4}t^{-1/4}= \dfrac{3}{4} 3t^{3/4}\)

\(\displaystyle f'(t) = t^{-1/4}= t^{3/4}\)

\(\displaystyle f'(t) = \dfrac{1}{t^{4}} = t^{3/4}\) :confused: What now?
Click to expand...
.
 
\(\displaystyle f(t) = t^{3/4} - 3t^{1/4} \)

\(\displaystyle f'(t) = \dfrac{3}{4}t^{-1/4} - \dfrac{1}{4}3t^{-3/4}\)

\(\displaystyle f'(t) = \dfrac{3}{4}t^{-1/4} - \dfrac{3}{4}t^{-3/4}\)

\(\displaystyle \dfrac{3}{4}t^{-1/4} - \dfrac{3}{4}t^{-3/4} = 0\)

\(\displaystyle \dfrac{3}{4}t^{-1/4}= \dfrac{3}{4} 3t^{-3/4}\)

\(\displaystyle t^{-1/4}= t^{-3/4}\)

\(\displaystyle \dfrac{1}{t^{-1/4}}= t^{-3/4}\)

\(\displaystyle 1 = \dfrac{t^{3/4}}{t^{1/4}}\)
 
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Jason76 said:
\(\displaystyle f(t) = t^{3/4} - 3t^{1/4} \)

\(\displaystyle f'(t) = \dfrac{3}{4}t^{-1/4} - \dfrac{1}{4}3t^{3/4}\) ....................Incorrect

\(\displaystyle f'(t) = \dfrac{3}{4}t^{-1/4} - \dfrac{3}{4}t^{3/4}\)

\(\displaystyle \dfrac{3}{4}t^{-1/4} - \dfrac{3}{4}t^{3/4} = 0\)

\(\displaystyle \dfrac{3}{4}t^{-1/4}= \dfrac{3}{4} 3t^{3/4}\)

\(\displaystyle t^{-1/4}= t^{3/4}\)

\(\displaystyle \dfrac{1}{t^{-1/4}}= t^{3/4}\)

\(\displaystyle 1 = \dfrac{t^{3/4}}{t^{1/4}}\)
Click to expand...
.
 
\(\displaystyle f(t) = t^{3/4} - 3t^{1/4} \)

\(\displaystyle f'(t) = \dfrac{3}{4}t^{-1/4} - \dfrac{1}{4}3t^{-3/4}\)

\(\displaystyle f'(t) = \dfrac{3}{4}t^{-1/4} - \dfrac{3}{4}t^{-3/4}\)

\(\displaystyle \dfrac{3}{4}t^{-1/4} - \dfrac{3}{4}t^{-3/4} = 0\)

\(\displaystyle \dfrac{3}{4}t^{-1/4}= \dfrac{3}{4} 3t^{-3/4}\)

\(\displaystyle t^{-1/4}= t^{-3/4}\)

\(\displaystyle \dfrac{1}{t^{1/4}}= t^{-3/4}\)

\(\displaystyle 1 =(t^{-3/4})(t^{1/4})\)

\(\displaystyle 1 = t^{-2/4}\)

\(\displaystyle 1 = t^{-1/2}\)

\(\displaystyle 1 = (t^{-1/2})^{-2}\)

\(\displaystyle 1^{-2} = t\) :confused: Please, What now?
 
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Jason76 said:
\(\displaystyle f(t) = t^{3/4} - 3t^{1/4} \)

\(\displaystyle f'(t) = \dfrac{3}{4}t^{-1/4} - \dfrac{1}{4}3t^{-3/4}\)

\(\displaystyle f'(t) = \dfrac{3}{4}t^{-1/4} - \dfrac{3}{4}t^{-3/4}\)

\(\displaystyle \dfrac{3}{4}t^{-1/4} - \dfrac{3}{4}t^{-3/4} = 0\)

\(\displaystyle \dfrac{3}{4}t^{-1/4}= \dfrac{3}{4} 3t^{-3/4}\)

\(\displaystyle t^{-1/4}= t^{-3/4}\)

\(\displaystyle \dfrac{1}{t^{1/4}}= t^{-3/4}\)

\(\displaystyle 1 =(t^{-3/4})(t^{1/4})\)

\(\displaystyle 1 = t^{-2/4}\)

\(\displaystyle 1 = t^{-1/2}\)

\(\displaystyle 1 = (t^{-1/2})^{-2}\)

\(\displaystyle 1^{-2} = t\) :confused: Please, What now?
Click to expand...

A little algebra.....

1-2 = 1
 
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Jason76 said:
\(\displaystyle f(t) = t^{3/4} - 3t^{1/4} \)

\(\displaystyle f'(t) = \dfrac{3}{4}t^{-1/4} - \dfrac{1}{4}3t^{-3/4}\) Correct

\(\displaystyle f'(t) = \dfrac{3}{4}t^{-1/4} - \dfrac{3}{4}t^{-3/4}\) Correct

\(\displaystyle \dfrac{3}{4}t^{-1/4} - \dfrac{3}{4}t^{-3/4} = 0\) OK

The line above is not generally true, but you are trying to find where t makes f'(t) = 0, where it is true.

Nevertheless, I'd prefer to say \(\displaystyle f'(t) = 0 \implies \dfrac{3}{4}t^{-1/4} - \dfrac{3}{4}t^{-3/4} = 0.\)

\(\displaystyle \dfrac{3}{4}t^{-1/4}= \dfrac{3}{4} t^{-3/4}\) Correct

\(\displaystyle t^{-1/4}= t^{-3/4}\) Correct

\(\displaystyle \dfrac{1}{t^{1/4}}= t^{-3/4}\) Correct, but you reduce likelihood of error if you eliminate negative exponents. See below.

\(\displaystyle 1 =(t^{-3/4})(t^{1/4})\) Correct

\(\displaystyle 1 = t^{-2/4}\) Correct

\(\displaystyle 1 = t^{-1/2}\) Correct

\(\displaystyle 1 = (t^{-1/2})^{-2}\) Correct

\(\displaystyle 1^{-2} = t\) :confused: Please, What now? When you raise 1 to a power, what do you get?
Click to expand...

\(\displaystyle t^{-(1/4)}= t^{-(3/4)} \implies\)

\(\displaystyle \dfrac{t^{-(1/4)}}{t^{-(3/4)}} = \dfrac{t^{-(3/4)}}{t^{-(3/4)}}\implies\)

\(\displaystyle t^{\{-(1/4) + (3/4)\}} = 1 \implies\)

\(\displaystyle t^{(2/4)} = 1 \implies\)

\(\displaystyle t^{(1/2)} = 1 \implies\)

\(\displaystyle \left(t^{(1/2)}\right)^2 = 1^2 \implies\)

\(\displaystyle t = 1.\)
 
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rewritng the derivative

\(\displaystyle \dfrac{3}{4\sqrt[4]{t}}\) - \(\displaystyle \dfrac{3}{4\sqrt[4]{t^{3}}} = 0\)

\(\displaystyle \dfrac{3}{4\sqrt[4]{t}} = \dfrac{3}{4\sqrt[4]{t^{3}}} \)

\(\displaystyle 12\sqrt[4]{t} = 12\sqrt[4]{t^{3}}\)

\(\displaystyle \sqrt[4]{t} = \sqrt[4]{t^{3}}\):confused:
 
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Jason76 said:
rewritng the derivative

\(\displaystyle \dfrac{3}{4\sqrt[4]{t}}\) - \(\displaystyle \dfrac{3}{4\sqrt[4]{t^{3}}} = 0\)

\(\displaystyle \dfrac{3}{4\sqrt[4]{t}} = \dfrac{3}{4\sqrt[4]{t^{3}}} \)

\(\displaystyle 12\sqrt[4]{t} = 12\sqrt[4]{t^{3}}\)

\(\displaystyle \sqrt[4]{t} = \sqrt[4]{t^{3}}\):confused:
Click to expand...

What is your confusion?
 
Jason76 said:
rewritng the derivative

\(\displaystyle \dfrac{3}{4\sqrt[4]{t}}\) - \(\displaystyle \dfrac{3}{4\sqrt[4]{t^{3}}} = 0\)

\(\displaystyle \dfrac{3}{4\sqrt[4]{t}} = \dfrac{3}{4\sqrt[4]{t^{3}}} \)

\(\displaystyle 12\sqrt[4]{t} = 12\sqrt[4]{t^{3}}\)

\(\displaystyle \sqrt[4]{t} = \sqrt[4]{t^{3}}\):confused:
Click to expand...
For goodness sake Jason, this is basic algebra.

\(\displaystyle \sqrt[4]{t} = \sqrt[4]{t^{3}} \implies\)

\(\displaystyle \left(\sqrt[4]{t}\right)^4 = \left(\sqrt[4]{t^{3}}\right)^4 \implies\)

\(\displaystyle t = t^3 \implies\)

\(\displaystyle 1 = t^2 \implies\)

\(\displaystyle t = 1.\) Minus 1 is not a possible answer because, by the initial conditions of the problem, t > 0.
 
JeffM said:
For goodness sake Jason, this is basic algebra.

\(\displaystyle \sqrt[4]{t} = \sqrt[4]{t^{3}} \implies\)

\(\displaystyle \left(\sqrt[4]{t}\right)^4 = \left(\sqrt[4]{t^{3}}\right)^4 \implies\)

* \(\displaystyle t = t^3 \implies\)

\(\displaystyle 1 = t^2 \implies \ \ \ \ \ \)X

\(\displaystyle t = 1.\) Minus 1 is not a possible answer because,

by the initial conditions of the problem, t > 0. \(\displaystyle \ \ \) The original post was edited by Jason76.
If "t > 0" was there to begin with, let me know please.
Click to expand...

We are looking for (first order) critical numbers, correct?

Before looking at the limitations on t, we have from * in the quote box above that:


\(\displaystyle t = t^3 \ \implies\)

\(\displaystyle t^3 - t = 0 \ \implies\)

\(\displaystyle t(t^2 - 1) = 0 \ \implies \)

\(\displaystyle t(t - 1)(t + 1) = 0 \ \implies\)

\(\displaystyle t = 0, \ \ t = 1, \ \ or \ \ t = -1 \ \ \ \ (potentially)\)


t = -1 isn't in the the domain of the function, and there is not a vertical tangent there.

t = 0 is in the domain of the function, and there is a vertical tangent (the derivative is undefined) there.

t = 1 is in the domain of the function, and there is some type of extrema (the derivative equals zero) there.


The (first order) critical numbers are t = 0 and t = 1.
 
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lookagain said:
We are looking for (first order) critical numbers, correct?

Before looking at the limitations on t, we have from * in the quote box above that:


\(\displaystyle t = t^3 \ \implies\)

\(\displaystyle t^3 - t = 0 \ \implies\)

\(\displaystyle t(t^2 - 1) = 0 \ \implies \)

\(\displaystyle t(t - 1)(t + 1) = 0 \ \implies\)

\(\displaystyle t = 0, \ \ t = 1, \ \ or \ \ t = -1 \ \ \ \ (potentially)\)


t = -1 isn't in the the domain of the function, and there is not a vertical tangent there.

t = 0 is in the domain of the function, and there is a vertical tangent (the derivative is undefined) there.

t = 1 is in the domain of the function, and there is some type of extrema (the derivative equals zero) there.


The (first order) critical numbers are t = 0 and t = 1.
Click to expand...
Lookagain

You may well be correct. Jason did not say what the problem was exactly. The way he attacked the problem implied to me that he was to find "the" local extremum. In particular he asked about the "single" value of t. I may have inferred what was not intentionally implied. If the problem was to find first order critical values of the underlying function, then I don't understand where the concept of t's "single value" came from. Although 0 is in the domain of the underlying function, it is not in the domain of the derivative.

The fact is, whenever a problem is paraphrased or implied, it's always possible to go astray when answering. But thanks for pointing out that my inference may have been incorrect.
 
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