problem with derivative

**furor celtica** · August 2nd 2010, 08:41 PM

ok so for the function f(x) = (x^(3/5))(x^(2)-13) (the many parentheses are to make it clear) i am supposed to find f'(x) and the intervals in which f(x) is decreasing and increasing
i found f'(x) = (13/5)(x^(-2/5))(x^(2)-3)
i think this is correct
but then to find the increasing and decreasing intervals i got it wrong
i started with the increasing interval and got x>sqrt3 and -sqrt3>x>0, i keep doing it over and over but i dont understand where its wrong

**Math Major** · August 2nd 2010, 08:55 PM

I disagree with your derivative.

$f(x) = (x^{\frac{3}{5}})(x^2 - 13)$

Using the product rule to differentiate,

$f'(x) = \frac{d}{dx}[(x^{\frac{3}{5}})] (x^2 -13) + (x^{\frac{3}{5}})\frac{d}{dx}(x^2 - 13)$

Applying the power rule,

$f'(x) = \frac{3}{5}(x^{\frac{-2}{5}})(x^2 - 13) + (x^{\frac{3}{5}})(2x)$

Combining terms,

$f'(x) = \frac{3}{5}(x^{\frac{8}{5}} - 13x^{\frac{-2}{5}}) + 2x^{\frac{8}{5}}$

Now, solve for the intervals where $f'(x) > 0$ and $f'(x) < 0$ to find out where it's increasing and decreasing.

**furor celtica** · August 2nd 2010, 10:09 PM

my derivative is right, man
that wasnt the question

**Math Major** · August 2nd 2010, 10:22 PM

Edit: Wow, I'm bad, sorry.

**earboth** · August 2nd 2010, 10:29 PM

Originally Posted by furor celtica

ok so for the function f(x) = (x^(3/5))(x^(2)-13) (the many parentheses are to make it clear) i am supposed to find f'(x) and the intervals in which f(x) is decreasing and increasing <=== impossible

i found f'(x) = (13/5)(x^(-2/5))(x^(2)-3) <=== OK
i think this is correct
but then to find the increasing and decreasing intervals i got it wrong
i started with the increasing interval and got x>sqrt3 <=== OK
and -sqrt3>x>0, <=== impossible: Here you state that $-\sqrt{3} > 0$
i keep doing it over and over but i dont understand where its wrong

Unfortunately you didn't post your working so I can't tell where and what you are doing wrong. I guess that you have made a sign error in your calculations because the correct answer is:

$0<x<\sqrt{3}~\vee~x>\sqrt{3}$

**furor celtica** · August 2nd 2010, 11:22 PM

earboth: by increasing AND decreasing i don't mean both at once, i mean i have to find the places where it is increasing and the places where it is decreasing, i.e. find where it is increasing and you will find where it is decreasing (the rest of the curve where it is not increasing)
sorry about the sign error, of course i meant -sqrt3<x<0
i started by finding where f(x) was increasing i.e. where f'(x) >0
can you just show me your calculations cos i dont know how to use latex and it will take me very long
i just used both parts of f'(x) in the way that they myust either both be positive or both negative for f'(x) to be positive
then for each case i compiled the inequalities to find where they coincided
math major: no problem bro

**earboth** · August 3rd 2010, 12:36 AM

Originally Posted by furor celtica

earboth: by increasing AND decreasing i don't mean both at once, i mean i have to find the places where it is increasing and the places where it is decreasing, i.e. find where it is increasing and you will find where it is decreasing (the rest of the curve where it is not increasing)
sorry about the sign error, of course i meant -sqrt3<x<0
i started by finding where f(x) was increasing i.e. where f'(x) >0
can you just show me your calculations cos i dont know how to use latex and it will take me very long
i just used both parts of f'(x) in the way that they myust either both be positive or both negative for f'(x) to be positive
then for each case i compiled the inequalities to find where they coincided
math major: no problem bro

1. You possibly noticed that my answer wasn't exact enough ... sorry!

2. Here are my calculations:

$f'(x)=\dfrac{13}5 \cdot \dfrac{x^2-3}{\sqrt[5]{x^2}}~,~x > 0$

f is increasing if f'(x) > 0.

A quotient is positive, that means it is greater than zero, if numerator and denominator have the same sign. Since the denominator is positive for all $x \ne 0$ the numerator must be greater than zero too:

$x^2 - 3 > 0 \implies\ \underbrace{x < -\sqrt{3}}_{not \ allowed}~\vee~ x > \sqrt{3}$

f is decreasing if f'(c) < 0.

A quotient is negative, that means it is smaller than zero, if numerator and denominator have different signs. Since the denominator is positive for all $x \ne 0$ the numerator must be smaller than zero:

$x^2 - 3 < 0 \implies\ -\sqrt{3} < x < \sqrt{3}$

but $-\sqrt{3}<x\leq 0$ is not defined and therefore f is decreasing for $0<x0\sqrt{3}$

EDIT: How about making a sketch of the graph of f to get an overview of the behaviour of f?

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