## Schrodinger 2 years ago Followup to a derivatives question earlier: I'm told to find all critical points and specify which extrema are maxima or minima. I'm having trouble just chunking through the numbers. (Problem below).

1. stamp

i am not a doctor but i will see what i can help you

2. Schrodinger

$y = x ^{2/3}(x+2)$$y \prime = \frac{ 5x+4 }{ 3\sqrt[3]{x} }$All critical points should occur at an endpoint or at y' = 0 or y' being undefined by the definition of a critical point. The answers that satisfy the definition are $x = 0, x = -\frac{ 4 }{ 5 }.$Plugging the x values back into y itself, I am supposed to get$y = 0, y =\frac{ 12 }{ 25 }10^{1/3}$I have no idea how that second one is gotten. First one, makes sense, everything is cancelled out by being multiplied by zero. Second one, you have $(-\frac{ 4 }{ 5 })^{2/3}(-\frac{ 4 }{ 5 }+2)$How the hell are they ending up with that neat, fractional answer with that last bit?

3. Schrodinger

(Oh, and pardon me, i'm just horrible at algebra. Always have, will be marginally better with time. Eventually. Maybe.)

4. stamp

y'(0) does not exist 5x + 4 = 0 when x = -4/5 so y'(-4/5) = 0 y'(-4/5) is your critical point

5. Schrodinger

y'(0) is still a critical point. It's undefined, but it doesn't not exist. And by the definition of a critical point [y'(0) is undefined, an end point within a defined interval or at a point where y' = 0], it is a critical point. My book lists it as an answer word for word, and my book is definitely right.

6. stamp

Well if x = 0 and x = -4/5 are you critical points, find f(0) and f(-4/5)

7. stamp

f(0) = 0

8. Schrodinger

I found them above. I'm just having trouble numerically working out the second one.

9. stamp

something is amiss

10. Schrodinger

What's amiss? I'm just bad with fractions being raised to fractional exponents less than one, especially when the base is also less than one, lol. I just don't understand how they're getting a clean, fractional answer from it. Whenever I take the cube root of those fractions I just get an irrational slew of numbers, not a clean-cut, dandy little fraction.

11. phi

$(-\frac{ 4 }{ 5 })^{2/3}(-\frac{ 4 }{ 5 }+2)$ $(\frac{ 4^2 }{ 5^2 })^{1/3}(\frac{ 6 }{ 5 })$ $(\frac{ 2\cdot 2^3 }{ 5^2 })^{1/3}(\frac{ 6 }{ 5 })$ $(\frac{ 5\cdot 2\cdot 2^3 }{ 5 \cdot 5^2 })^{1/3}(\frac{ 6 }{ 5 })$ $(\frac{ 10\cdot 2^3 }{ 5^3 })^{1/3}(\frac{ 6 }{ 5 })$ $10^{1/3}(\frac{ 2 }{ 5 })(\frac{ 6 }{ 5 })$ $10^{1/3}\frac{ 12 }{ 25 }$

12. Schrodinger

@phi , Line 4, did you just multiply the left portion by 5/5?

13. phi

yes, because it would be nice to have 5^3 in the bottom, so we can take its cube root.

14. Schrodinger

Makes sense, just making sure. Thanks so much, dude, this helps a lot.