Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

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).

See more answers at
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly


Get your free account and access expert answers to this and thousands of other questions

i am not a doctor but i will see what i can help you
\[y = x ^{2/3}(x+2)\]\[y \prime = \frac{ 5x+4 }{ 3\sqrt[3]{x} }\]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?
(Oh, and pardon me, i'm just horrible at algebra. Always have, will be marginally better with time. Eventually. Maybe.)

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

y'(0) does not exist 5x + 4 = 0 when x = -4/5 so y'(-4/5) = 0 y'(-4/5) is your critical point
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.
Well if x = 0 and x = -4/5 are you critical points, find f(0) and f(-4/5)
f(0) = 0
I found them above. I'm just having trouble numerically working out the second one.
something is amiss
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.
  • phi
\[ (-\frac{ 4 }{ 5 })^{2/3}(-\frac{ 4 }{ 5 }+2) \] \[ (\frac{ 4^2 }{ 5^2 })^{1/3}(\frac{ 6 }{ 5 }) \] \[ (\frac{ 2\cdot 2^3 }{ 5^2 })^{1/3}(\frac{ 6 }{ 5 }) \] \[ (\frac{ 5\cdot 2\cdot 2^3 }{ 5 \cdot 5^2 })^{1/3}(\frac{ 6 }{ 5 }) \] \[ (\frac{ 10\cdot 2^3 }{ 5^3 })^{1/3}(\frac{ 6 }{ 5 }) \] \[ 10^{1/3}(\frac{ 2 }{ 5 })(\frac{ 6 }{ 5 }) \] \[ 10^{1/3}\frac{ 12 }{ 25 }\]
@phi , Line 4, did you just multiply the left portion by 5/5?
  • phi
yes, because it would be nice to have 5^3 in the bottom, so we can take its cube root.
Makes sense, just making sure. Thanks so much, dude, this helps a lot.

Not the answer you are looking for?

Search for more explanations.

Ask your own question