# critical points

• Nov 14th 2006, 10:42 PM
m777
critical points
please try to solve these questions.
• Nov 15th 2006, 10:31 AM
Soroban
Hello, m777!

Quote:

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

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

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

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

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

Quote:

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

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

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

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

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

Therefore:

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

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