# Thread: [SOLVED] local min and max

1. ## [SOLVED] local min and max

let f(x)= (x-1)^(4/5) x find the local minimum and maximum

2. Originally Posted by cole123
let f(x)= (x-1)^(4/5) x find the local minimum and maximum
$\displaystyle f(x) = (x - 1)^{\frac{4}{5}}x$

To find local maxima and minima, you need to find the derivative, set it equal to 0 and solve for $\displaystyle x$.

So $\displaystyle f'(x) = (x - 1)^{\frac{4}{5}}\frac{d}{dx}(x) + x\frac{d}{dx}[(x - 1)^{\frac{4}{5}}$

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

Can you go from here?

3. I am on the next step :

I have (4/5)x + (x-1)= 0

I know there is both a min and a max just dont know what to do from here

4. Originally Posted by cole123
I am on the next step :

I have (4/5)x + (x-1)= 0

I know there is both a min and a max just dont know what to do from here
No.

It's $\displaystyle (x - 1)^{\frac{4}{5}} + \frac{4}{5}x(x - 1)^{-\frac{1}{5}} = 0$.

5. I guess I dont then!!

6. Can you help me?

7. Prove It, cole123 was heading in the right direction. After you multiply $\displaystyle (x-1)^{\frac{4}{5}}+ \frac{4}{5}x(x-1)^{-\frac{1}{5}}= 0$ by $\displaystyle (x-1)^{1}{5}$, you are left with $\displaystyle (x-1)+ \frac{4}{5}x= 0$.

cole123, solve that: $\displaystyle x+ \frac{4}{5}x= 1$.

Remember that the local max and min must be at a point where the derivative is 0, or where there is NO derivative.

8. Originally Posted by cole123
I am on the next step :

I have (4/5)x + (x-1)= 0

I know there is both a min and a max just dont know what to do from here

I did that and someone said it was wrong

9. so is x = 5/9 my only answer? Ive only answered max/min for parabolas. and I believe there could be another answer?!

10. Originally Posted by HallsofIvy
Prove It, cole123 was heading in the right direction. After you multiply $\displaystyle (x-1)^{\frac{4}{5}}+ \frac{4}{5}x(x-1)^{-\frac{1}{5}}= 0$ by $\displaystyle (x-1)^{1}{5}$, you are left with $\displaystyle (x-1)+ \frac{4}{5}x= 0$.

cole123, solve that: $\displaystyle x+ \frac{4}{5}x= 1$.

Remember that the local max and min must be at a point where the derivative is 0, or where there is NO derivative.
OK point taken.

Except when you multiply both sides by $\displaystyle (x - 1)^{\frac{1}{5}}$ the equation actually becomes

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

$\displaystyle x^4 - 4x^3 + 6x^2 - 4x + 1 + \frac{4}{5}x = 0$.

Now simplify and solve for $\displaystyle x$.

11. Well, whoever told you that was wrong!

I suspect they graphed that on a calculator and saw no graph for x< 1.

Many perfectly good calculators (like my TI 89) will not handle odd roots of negative numbers correctly because they do them using a logarithm which is not defined for $\displaystyle x\le 0$.

If you graph y1= (x-1)^(4/5)*x on a TI calculator, for example, you will see a graph that begins at (1, 0) and rises to the right.

If you also graph y2= (1- x)^(4/5)*x, you will see the rest of the graph which has maximum at x= 5/9.

12. Originally Posted by Prove It
OK point taken.

Except when you multiply both sides by $\displaystyle (x - 1)^{\frac{1}{5}}$ the equation actually becomes

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

$\displaystyle x^4 - 4x^3 + 6x^2 - 4x + 1 + \frac{4}{5}x = 0$.

Now simplify and solve for $\displaystyle x$.
No! $\displaystyle (x- 1)^{\frac{4}{5}}(x-1)^{\frac{1}{5}}= (x- 1)^1$, not $\displaystyle (x- 1)^4$.