Critical Number Problem

[Jason76]

Find the critical numbers:

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

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

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

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

\(\displaystyle h'(t) = \dfrac{3}{4} t^{-1/4} - \dfrac{3}{4} t^{-3/4}\) = 0 Next move. How can I solve for t in this situation?

 

[Deleted member 4993]

Find the critical numbers:

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

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

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

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

\(\displaystyle h'(t) = \dfrac{3}{4} t^{-1/4} - \dfrac{3}{4} t^{-3/4}\) = 0 Next move. How can I solve for t in this situation? ...... simple algebra

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

let u = t^-1/4

3/4 * u - 3/4 * u³ = 0

3/4 * u * (1 - u²) = 0 → u = 0 and u = ± 1 ................ and continue........

 

[Jason76]

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

let u = t^-1/4

3/4 * u - 3/4 * u³ = 0

3/4 * u * (1 - u²) = 0 → u = 0 and u = ± 1 ................ and continue........

3/4 * u - 3/4 * u³ = 0

Isn't this a typo. Shouldn't it be 3/4 * u - 3/4 * u^-3/4 = 0 ?

 

[srmichael]

Isn't this a typo. Shouldn't it be 3/4 * u - 3/4 * u^-3/4 = 0 ?

No.

S Khan let \(\displaystyle u=t^{\frac{-1}{4}}\) and \(\displaystyle u^3=(t^{\frac{-1}{4}})^3=t^{\frac{-3}{4}}\)

 

[Jason76]

Find the critical numbers:

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

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

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

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

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

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

\(\displaystyle h'(t) = t^{-3/4} = t^{-1/4} = t^{-4/4} = t^{-1}\) :?:

 

[lookagain]

First, to understand what you are doing, it is not generally true that h'(t) = 0.


You are trying to find the values of t where h'(t) = 0 \(\displaystyle \ \ \) and where h'(t) is undefined.


so

\(\displaystyle h'(t) = 0 \implies \)


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


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


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


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


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

\(\displaystyle 1 = t^{-(1/2)} \ \implies \ \ \ \) No, it does not imply the line after this line.

\(\displaystyle 1 = \dfrac{1}{t^2}\)


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


\(\displaystyle t = \pm 1.\)

In fact, \(\displaystyle \ \ 1 = t^{-(1/2)} \ \implies \ \ \ \)

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



t = 0 is also a critical number as Subhotosh Khan showed/included. \(\displaystyle \ \ \) The first derivative is undefined at t = 0.

 

[Jason76]

Find the critical numbers:

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

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

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

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

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

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

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

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

\(\displaystyle t^{1/2} = 0 \) :?:

 

[JeffM]

In fact, \(\displaystyle \ \ 1 = t^{-(1/2)} \ \implies \ \ \ \)

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



t = 0 is also a critical number as Subhotosh Khan showed/included. \(\displaystyle \ \ \) The first derivative is undefined at t = 0.

You are of course right, lookagain. Something must have distracted me while doing the post, which I have deleted. Thanks.