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
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.
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
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?
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