# differentiate and find maximum and minimum turning point

• October 1st 2011, 08:42 AM
salma2011
differentiate and find maximum and minimum turning point
2(x^2/3) + (x^4/5) +11

i differnetiated this to get
4/3(x^-1/3) + 4/5(x^-1/5)

i know then, that i have to set the differentiated formula = 0. Please help with what i have to do next.

Thanks
• October 1st 2011, 08:46 AM
Siron
Re: differentiate and find maximum and minimum turning point
The first derivative is indeed:
$f'(x)=\frac{4}{3}x^{\frac{-1}{3}}+\frac{4}{5}x^{\frac{-1}{5}}$
To find the minumum/maximum then you have to solve $f'(x)=0$. What do you get? Afterwards make a sign table to determine where the function increases/decreases.
• October 1st 2011, 08:59 AM
salma2011
Re: differentiate and find maximum and minimum turning point
I can't remember how to solve it. i know im missing something simple. so can anyone help me solve this please. thank you
• October 1st 2011, 10:08 AM
e^(i*pi)
Re: differentiate and find maximum and minimum turning point
Quote:

Originally Posted by salma2011
I can't remember how to solve it. i know im missing something simple. so can anyone help me solve this please. thank you

I suggest finding the lowest common denominators of 3 and 5 and change your exponents into that form:

$f'(x) = \dfrac{4}{3}x^{-5/15} + \dfrac{4}{5}x^{-3/15} = 0$

You can then factor out $4x^{-1/15}$ as a common factor

$4x^{-3/15}\left(\dfrac{x^2}{3} + \dfrac{1}{5}\right) = 0$

Can you solve now?

To test whether you have a maximum or minimum check the sign of $f''(x)$.

• If $f''(x) < 0$ you have a maximum
• If $f''(x) > 0$ you have a minimum
• If $f''(x) = 0$ you have a point of inflection
• October 1st 2011, 10:13 AM
Siron
Re: differentiate and find maximum and minimum turning point
There're probably different ways to solve the equation, but if you have to solve:
$0=\frac{4}{3}x^{\frac{-1}{3}}+\frac{4}{5}x^{\frac{-1}{5}}$
$\Leftrightarrow 4x^{\frac{-1}{3}}\left(\frac{1}{3}+\frac{1}{5}x^{\frac{2}{15} }\right)=0$
$\Leftrightarrox x^{\frac{-1}{3}}=0 \ \mbox{or} \ x^{\frac{2}{15}}=\frac{-5}{3}$

This means there're no solutions, because there's no $x$ value wherefore:
$x^{\frac{-1}{3}}=0$ or $x^{\frac{2}{15}}=\frac{-5}{3}$.