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)
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)
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
Hello, turillian!
I think I have the first one . . .
First, we note that: .$\displaystyle 8 - x^3\:>\:0\quad\Rightarrow\quad x < 2$1) Find the intervals of concavity for: .$\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: .$\displaystyle (\text{-}\infty,\:\text{-}\sqrt[3]{16}) \:\cup\:(0,\:2)$
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
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.
correct. though your method does not clearly show how you got $\displaystyle \pm e^{-1} $
for this problem we need implicit differentiation, are you familiar with that?
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 $\displaystyle \frac {dy}{dx}$