# critical points

• Nov 14th 2006, 09:42 PM
m777
critical points
please try to solve these questions.
• Nov 15th 2006, 09:31 AM
Soroban
Hello, m777!

Quote:

2) Find $\displaystyle \frac{dy}{dx}$ if $\displaystyle x^3 + xy + y^3\:=\:6$

Differentiate implicitly: .$\displaystyle 3x^2 + x\left(\frac{dy}{dx}\right) + y + 3y^2\left(\frac{dy}{dx}\right) \;= \;0$

Then: .$\displaystyle x\left(\frac{dy}{dx}\right) + 3y^2\left(\frac{dy}{dx}\right) \;= \;-3x^2 - y$

Factor: .$\displaystyle \left(x^2 + 3y^2\right)\!\cdot\!\frac{dy}{dx} \;= \;-(3x^2 + y)$

Therefore: .$\displaystyle \frac{dy}{dx} \;= \;-\frac{3x^2+y}{x+3y^2}$

Quote:

3) a) What are the critical points of $\displaystyle f(x)$ ?
b) On what interval(s) is $\displaystyle f(x)$ increasing or decreasing?
. . . . $\displaystyle f'(x) \:=\:(x-1)^2(x+2)$

(a) Critical points occur where $\displaystyle f'(x) = 0$
. . .This happens when: .$\displaystyle x\:=\:\text{-}2,\,1$

To the left of $\displaystyle x = \text{-}2$, we have: .$\displaystyle f'(\text{-}3) \:=\:(\text{-}4)^2(-1) \:=\:\text{-}16$ .Decreasing: $\displaystyle \searrow$

Between $\displaystyle x = \text{-}2$ and $\displaystyle x = 1$, we have: .$\displaystyle f'(0) \:=\:(-1)^2(2^2) \:=\:+4$ .Increasing: $\displaystyle \nearrow$

To the right of $\displaystyle x = 1$, we have: .$\displaystyle f'(2)\:=\:(1^2)(4^2) \:=\:+16$ .Increasing: $\displaystyle \nearrow$

Therefore:

. . $\displaystyle f(x)$ is decreasing on $\displaystyle (\text{-}\infty,\,\text{-}2)$

. . $\displaystyle f(x)$ is increasing on $\displaystyle (\text{-}2,\,1) \cup (1,\,\infty)$