These two in particular are really thumping around inside my head going nowhere in particular, any help would be greatly appreciated.

Find the intervals of concavity for: y=ln sqrt(8-x^3)

Find the regions of increase and decrease for: y=xln(x^2)

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- May 27th 2007, 12:41 PMturillian@gmail.comExponents and Logarithms Calculus questions.
These two in particular are really thumping around inside my head going nowhere in particular, any help would be greatly appreciated.

Find the intervals of concavity for: y=ln sqrt(8-x^3)

Find the regions of increase and decrease for: y=xln(x^2) - May 27th 2007, 01:01 PMJhevon
Some definitions:

A function is said to be**concave up**on a certain interval if its second derivative is positive on that interval, it is said to be**concave down**on the interval if its second derivative is negative on the interval

A function is said to be**increasing**on an interval if its first derivative is positive on the interval, it is said to be**decreasing**on an interval if its first derivative is negative on the interval

so to do the first question, find the second derivative $\displaystyle y''$, then solve $\displaystyle y'' < 0 $ to find where it's concave down and solve $\displaystyle y'' > 0 $ to find where it is concave up

for the second question, find $\displaystyle y'$, solve $\displaystyle y' > 0 $ to find where the function is increasing and solve $\displaystyle y' < 0 $ to find where the function is decreasing

do you understand? if not, ask questions, if yes, do the problems and post your solutions so we can check it

EDIT: Soroban seems to be posting a solution as we speak, so you can check your work against his - May 27th 2007, 02:04 PMSoroban
Hello, turillian!

I*think*I have the first one . . .

Quote:

1) Find the intervals of concavity for: .$\displaystyle y\:=\:\ln\left(\sqrt{8-x^3}\right)$

We have: .$\displaystyle y \:=\:\ln(8-x^3)^{\frac{1}{2}} \:=\:\frac{1}{2}\ln(8-x^3)$

Differentiate: .$\displaystyle y' \;=\;\frac{1}{2}\cdot\frac{-3x^2}{8-x^3} \;=\;\frac{3}{2}\cdot\frac{x^2}{x^3-8}$

Differentiate: .$\displaystyle y'' \;=\;\frac{3}{2}\left[\frac{(x^3-8)2x - x^2(3x^2)}{(x^3-8)^2}\right]\;=\;-\frac{3}{2}x\cdot\frac{x^3+16}{(x^3-8)^2}$

We see that $\displaystyle y'' = 0$ .when $\displaystyle x = 0$ .and .$\displaystyle x = -\sqrt[3]{16} \approx -2.52$

We have two critical values: .$\displaystyle -\sqrt[3]{16},\:0$

. . and three intervals to test.

On $\displaystyle (\text{-}\infty,\:\text{-}\sqrt[3]{16})$, try $\displaystyle x = \text{-}3$

$\displaystyle y''(\text{-}3)\;=\;\left(-\frac{3}{2}\right)(-3)\cdot\frac{-27+16}{(-27-8)^2} \:=\:\left(+\frac{9}{2}\right)\cdot\frac{(-)}{(+)} \;=\;(-)$ . . . concave down

On $\displaystyle (\text{-}\sqrt[3]{16},\:0)$, try $\displaystyle x = \text{-}1$

$\displaystyle y''(\text{-}1) \;=\;\left(-\frac{3}{2}\right)(-1)\cdot\frac{-1 + 16}{(-1-8)^2} \;=\;\left(+\frac{3}{2}\right)\cdot\frac{(+)}{(+)} \;=\;(+)$ . . . concave up

On $\displaystyle (0,\:2)$, try $\displaystyle x = 1$

$\displaystyle y''(1)\;=\;\left(-\frac{3}{2}\right)(1)\cdot\frac{1 + 16}{(1-8)^2}\;=\;\left(-\frac{3}{2}\right)\cdot\frac{(+)}{(+)} \;=\;(-)$ . . . concave down

Therefore, the function is concave up on: .$\displaystyle (\text{-}\sqrt[3]{16},\:0)$

. . and concave down on: .$\displaystyle (\text{-}\infty,\:\text{-}\sqrt[3]{16}) \:\cup\:(0,\:2)$

- May 27th 2007, 02:40 PMJhevon
Soroban seems to have neglected this one for some reason, so let me start you off.

$\displaystyle y = x \ln \left( x^2 \right)$

$\displaystyle \Rightarrow y = 2x \ln(x)$

$\displaystyle \Rightarrow y ' = 2 \ln(x) + 2 = \ln \left( x^2 \right) + 2$ .........By the product rule

Now find for what values of $\displaystyle x$ is $\displaystyle y'$ positive or negative

Note: We did not have to simplify the ln before taking the derivative, but i figured it would eb easier to see that way - May 27th 2007, 08:14 PMturillian@gmail.com
Okay, I ended up with the following for "Find the regions of increase and decrease for: y=xln(x^2)" :

critical values happen when ln(x^2)+2 = 0.

ln(x^2)=-2

2ln(x)=-2

ln(x)=-1

x=(+/-)e^-1

so the critical values are plus or minus (e^-1).

Then from there the intervals are straight forward and easy to test.

Thanks a lot, I kind of just forgot the general approach to solving these.

I have another problem, I don't know where to start.

Find dy/dx of e^(xy) = x^3 + 2x + xy

Thank you all, thank you. - May 27th 2007, 08:24 PMJhevon
correct. though your method does not clearly show how you got $\displaystyle \pm e^{-1} $

Quote:

Thanks a lot, I kind of just forgot the general approach to solving these.

I have another problem, I don't know where to start.

Find dy/dx of e^(xy) = x^3 + 2x + xy

Thank you all, thank you.

we differentiate normally, but whenever we differentiate a y term, we attach a dy/dx to it (since we took the derivative of y with respect to x). Note that we also attach something when we differentiate an x term, we attach dx/dx (the derivative of x with respect to x), but derivative notations can function as fractions and so dx/dx becomes 1 and we dont see it. whenever we have a term made up of x and y, we use the product rule and attach dy/dx when we differentiate the y part and so on (when differentiating with respect to y we treat x as a constant, when differentiating with respect to x we treat y as a constant).

$\displaystyle e^{xy} = x^3 + 2x + xy$

Differentiating implicitly we obtain:

$\displaystyle ye^{xy} + xe^{xy} \frac {dy}{dx} = 3x^2 + 2 + y + x \frac {dy}{dx}$

Now just solve for $\displaystyle \frac {dy}{dx}$ - May 28th 2007, 08:33 PMturillian@gmail.com
I'm in your debt, thanks a lot.

Good to know about the notation.