Finding Critical Numbers...

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jrujreport

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I want to find the critical numbers of the problem:
x3/5 * (x+3) = f(x)

so...


d/dx (x3/5 * (x+3)) = d/dx(x3/5)(x+3) + x3/5*d/dx(x+3)
= 3/5 x-2/5(x+3) + x3/5
= 3/5 x3/5 + 9/5 x-2/5 + x3/5
= 8/5 x3/5 + 9/5 x-2/5
Given this derivative, you just need to solve the equation:
8/5 x3/5 + 9/5 x-2/5 = 0

and I end up finding -1 as an answer, but Webwork says I am wrong.

What gives?

Thanks!
 
jrujreport said:
I want to find the critical numbers of the problem:
x3/5 * (x+3) = f(x)

so...


d/dx (x3/5 * (x+3)) = d/dx(x3/5)(x+3) + x3/5*d/dx(x+3)
= 3/5 x-2/5(x+3) + x3/5
= 3/5 x3/5 + 9/5 x-2/5 + x3/5
= 8/5 x3/5 + 9/5 x-2/5
Given this derivative, you just need to solve the equation:
8/5 x3/5 + 9/5 x-2/5 = 0

and I end up finding -1 as an answer, but Webwork says I am wrong.

What gives?

Thanks!
Click to expand...

Does x = -1 satisfy 8/5 x3/5 + 9/5 x-2/5 = 0 ?

Anyway:

8/5 x3/5 + 9/5 x-2/5 = 0

1/5 x-2/5 * (8x + 9) = 0

x = - 9/8

 
Common denominator method

Express the derivative as 8/5 x^3/5 + 9/(5x^(2/5))
With a common denominator of 5x^(2/5) you get
(8x+9)/(5x^(2/5))
So your critical numbers are x =-9/8 from the numerator (relative maximum) and x=0 from the denominator (relative minimum).

I hope this helps!
 
Cheezees said:
Express the derivative as 8/5 x^(3/5) + 9/(5x^(2/5))
With a common denominator of 5x^(2/5) you get
(8x+9)/(5x^(2/5))
So your critical numbers are x =-9/8 from the numerator (relative maximum) \(\displaystyle \ \ \ \ \)No, a relative minimum occurs there. *

and x=0 from the denominator (relative minimum).\(\displaystyle \ \ \ \ \)Neither a relative minimum nor a relative maximum occurs there.
At x = 0 there is an inflection point. **
Click to expand...

* For instance, if you test x = -2 and then x = -1 in the first derivative, the signs of he first derivative
change from negative to positive.


** If you test x-values close enough on either side of x = 0 in the second derivative, the signs of the
second derivative change. (The concavity changes.)
 
Last edited:
f(x) is not defined for x<0, over the reals
f(x) is a strictly increasing function
 
Last edited:
Bob Brown MSEE said:
f(x) is not defined for x<0, over the reals
f(x) is a strictly increasing function
Click to expand...

That is not true. The domain of \(\displaystyle \ f(x) \ = \ (x + 3)x^{3/5} \ \ is \ \ (-\infty, \infty).\)*

f(x) is not strictly increasing.



* This pretty much hinges on the fact that some base function, call it \(\displaystyle \ g(x) \ = \ x^{1/5} \ = \ \sqrt[5]{x}, \)

is defined for all real numbers.
 
Last edited:
lookagain said:
* For instance, if you test x = -2 and then x = -1 in the first derivative, the signs of he first derivative
change from negative to positive.


** If you test x-values close enough on either side of x = 0 in the second derivative, the signs of the
second derivative change. (The concavity changes.)
Click to expand...

YES, you are correct!!! I made an error with my signs! :)
 
Last edited by a moderator:
lookagain said:
That is not true. The domain of \(\displaystyle \ f(x) \ = \ (x + 3)x^{3/5} \ \ is \ \ (-\infty, \infty).\)*

f(x) is not strictly increasing.



* This pretty much hinges on the fact that some base function, call it \(\displaystyle \ g(x) \ = \ x^{1/5} \ = \ \sqrt[5]{x}, \)

is defined for all real numbers.
Click to expand...


Click
 
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