Question

find all x-coordinates of all the relative extreme & inflection points of f(x) = 1/3Px^3 + 2sqrt(P)x^2 - 5x +2. Where p is greater that 0 & is a constant. We were told not to substitut a number in for P.

I just am not sure if i am doing it right, I found first derivative of = px^2 + 4sqrt(p)x - 5 = and having a very hard time knowing if i did this right = Critical Points = (-4sqrt(p))/2p

Also for inflection points I took the second derivative and found it to = 2px+ 4sqrt(p) & the zero of that I found to be = -2/sqrt(p) = inflection points - am I correct at all on these or way off?

Answer 1

f(x) = 1/3Px^3 = 2sqrt(P)x^2 - 5x +2
Is the second equal sign supposed to be + ?

Answer 2

Yes, so sorry

Answer 3

Just fixed that

Answer 4

\[
f(x) = \frac 13Px^3 + 2\sqrt{P}x^2 - 5x +2 \\
f'(x) = Px^2 + 4\sqrt{P}x - 5 = 0 \\
x = \frac{-4\sqrt{P} \pm \sqrt{16P + 20P}}{2P} =
\frac{-4\sqrt{P} \pm 6\sqrt{P}}{2P} =
\frac{-5\sqrt{P}}{P}, ~~~\frac{\sqrt{P}}{P}\\
\]

Answer 5

Thank you! I was doing the quad equation and forgot to square the stinkin 4 as well and was coming up with weird answers. Is my critical point correct?

Answer 6

The two x values shown above are the critical points:\[
x = \frac{-5\sqrt{P}}{P}, ~~~x = \frac{\sqrt{P}}{P}\\
\]

Answer 7

Sorry, I meant inflection points

Answer 8

\[
f''(x) = 2Px + 4\sqrt{P} = 0 \\
x = -\frac{-4\sqrt{P}}{2P} = -2\frac{\sqrt{P}}{P}
\]

Answer 9

Can you simplify that to -2/sqrt(p)? or not?

Answer 10

Many textbooks and teachers don't like a radical in the denominator.
They ask you to rationalize the denominator.

Answer 11

Ok, but it is the same thing? Just making sure i understand that math, and my teacher is a little bizzar

Answer 12

They are identical.

Answer 13

Thanks so much for your help!!

Answer 14

They are identical. But some teachers and online test software mark the answer wrong if the denominator is not rationalized. So I will play it safe expressing the answers for both the critical points and the inflection point in the form I did with no radicals in the denominator.

You are welcome.