Question

Find the x-coordinates of any relative extrema and inflection point(s) for the function f(x) = 6x^(1/3) + 3x^(4/3). You must justify your answer using an analysis of f ′(x) and f ′′(x).

Answer 1

relative extremas can be determined by finding the first derivative, setting it equal to 0, then finding your critical points and testing where they change signs.

Answer 2

Well i know the derivative is 2x^-2/3 + 4x^1/3 but i forget how to solve for x here

Answer 3

\[\ f(x) = 6x^{\frac{1}{3}} + 3x^{\frac{4}{3}}\]\[\ f'(x) = 6 \cdot \frac{1}{3}x^{-\frac{2}{3}} + 3 \cdot \frac{4}{3} x^{\frac{1}{3}}\]\[\ f'(x) = 2x^{-\frac{2}{3}} + 4x^{\frac{1}{3}} = 0 \] Now you need to factor out the LCM

And good job getting the derivative :)

Answer 4

Between 2 and 4, 2 is the LCM, and between\(\ -\frac{2}{3} \&\ \frac{1}{3}\) , \(\ -\frac{2}{3}\) is smaller. right?

Answer 5

so the equation would no be 2x^-2/3(x + 2x^1/2) i presume

Answer 6

well, when you multiply \(\ -\frac{2}{3} \cdot \frac{1}{2}\) do you get \(\frac{1}{3}\)? :o

Answer 7

yea my mistake so 2x^-2/3(x + 2x^-1/2)

Answer 8

Yep! and now you set this equal to 0

Answer 9

\(\ \implies 2x^{-\frac{2}{3}} = 0 \ \ and\ \ x+ 2x^{-\frac{1}{2}} = 0\)

Answer 10

This is where i get confused because i forget how to solve these kinds of equations

Answer 11

Hmm... one sec.

Answer 12

You know.. we were going about the hard way of doing it :(

Answer 13

Sorry!! Let's try this again starting with this step.
\[\ f'(x) = 6 \cdot \frac{1}{3}x^{-\frac{2}{3}} + 3 \cdot \frac{4}{3} x^{\frac{1}{3}}\] from here, we can rewrite our powers as positive powers as well as reduce our function. \[\ \large f'(x) = \frac{2}{x^{\frac{2}{3}}} + 4x^{\frac{1}{3}}\]

Answer 14

Now we find a common denominator, \(\ x^{\frac{2}{3}}\)

Answer 15

So we'll end up with \[\ \large f'(x) = \frac{4x + 2}{x^{\frac{2}{3}}}\] :) From here...

Answer 16

Ohhhh!!! this makes much more sense

so the critical points are know x = 0, .5

Answer 17

We set \[\ \large x^{\frac{2}{3}} = 0 \implies x=0\] \[\ 4x+2 = 0 \implies x =\frac{1}{2} \]

Answer 18

Yesss you are right :)

Answer 19

and so those are your critical points of the relative extrema, and for the inflection points, we have to find the second derivative

Answer 20

we can use the product rule! All we have to do is rewrite our derivative in a multiplication form :) \[\ \large f'(x) = (4x+2)(x^{\frac{2}{3}})\] Then we just apply the product rue to this "\(\ f''g + g'f\)"

Answer 21

wouldn't it be the quotient rule?

Answer 22

I meant \[\ f'(x) = (4x+2)(x^{-\frac{2}{3}})\]

Answer 23

I see.

Answer 24

You can use the quotient rule if you'd like to leave your derivative as it is, then \(\ f = 4x+2 \ \ and \ \ g = x^{\frac{2}{3}}\) and you just apply the quotient rule, that is \(\ \frac{f'g - g'f}{g^2}\)

Answer 25

But I prefer the product rule since I don't like dealing with fractions :P

Answer 26

Oh, i typoed the product rule too D: Grr. It's f'g + g'f **

Answer 27

so f''(x) = 4x^-2/3 + -2/3x^-5/3(4x + 2)

Answer 28

Not the second derivative of f x_x

Answer 29

\[\ \large f'(x) = (4x+2)(x^{-\frac{2}{3}})\]\[\ f'g + g'f\]\[\ 4(x^{-\frac{2}{3}}) + \left(-\frac{2}{3}\right)x^{-\frac{5}{3}} \cdot (4x+2)\]

Answer 30

Solving this i'm realizing there is no easy way of going about finding your inflection points :(

Answer 31

I think i can figure it out from here and you have my sincerest thanks for helping me

Answer 32

And let's try the quotient rule with this.

Answer 33

If we have.... \[\ \large f'(x) = \frac{4x + 2}{x^{\frac{2}{3}}}\] Then the quotient rule will give us \[\frac{4(x^{\frac{2}{3}})-\left(\frac{2}{3}\right)x^{-\frac{1}{3}}(4x+2)}{\left(x^{\frac{2}{3}}\right)^2}\]

Answer 34

It will reduce down to \[\large -\frac{4}{3x^{\frac{5}{3}}} + 4\left( \frac{1}{3x^{\frac{2}{3}}}\right)\] which will further reduce to \[f''(x) =\frac{4(x-1)}{3x^{\frac{5}{3}}}\]

Answer 35

From here you just follow the same steps in finding your critical points, as we did earlier with the first derivative, and you'll find your inflection points

Answer 36

Good luck!