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

Printable View

- Oct 1st 2011, 08:42 AMsalma2011differentiate 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 - Oct 1st 2011, 08:46 AMSironRe: differentiate and find maximum and minimum turning point
The first derivative is indeed:

$\displaystyle f'(x)=\frac{4}{3}x^{\frac{-1}{3}}+\frac{4}{5}x^{\frac{-1}{5}}$

To find the minumum/maximum then you have to solve $\displaystyle f'(x)=0$. What do you get? Afterwards make a sign table to determine where the function increases/decreases. - Oct 1st 2011, 08:59 AMsalma2011Re: 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

- Oct 1st 2011, 10:08 AMe^(i*pi)Re: differentiate and find maximum and minimum turning point
I suggest finding the lowest common denominators of 3 and 5 and change your exponents into that form:

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

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

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

Can you solve now?

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

- If $\displaystyle f''(x) < 0$ you have a maximum
- If $\displaystyle f''(x) > 0 $ you have a minimum
- If $\displaystyle f''(x) = 0$ you have a point of inflection

- Oct 1st 2011, 10:13 AMSironRe: differentiate and find maximum and minimum turning point
There're probably different ways to solve the equation, but if you have to solve:

$\displaystyle 0=\frac{4}{3}x^{\frac{-1}{3}}+\frac{4}{5}x^{\frac{-1}{5}}$

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

$\displaystyle \Leftrightarrox x^{\frac{-1}{3}}=0 \ \mbox{or} \ x^{\frac{2}{15}}=\frac{-5}{3}$

This means there're no solutions, because there's no $\displaystyle x$ value wherefore:

$\displaystyle x^{\frac{-1}{3}}=0$ or $\displaystyle x^{\frac{2}{15}}=\frac{-5}{3}$.

So what's your conclusion? - Oct 1st 2011, 10:59 AMsalma2011Re: differentiate and find maximum and minimum turning point
i solved for x and got x = + or - square root of 3/5

(sorry don't know how to put mathematical functions onto the computer)

and then from the derivative i work out whether it's a maximum or a minimum.

thank you for all your help