Question

Find the critical numbers of the function.
g(x) = x^1/6 - x^5/6

Answer 1

\[g'(x)=\frac{1}{6}*x^{\frac{1}{6}-1}-\frac{5}{6}*x^{\frac{5}{6}-1}=\frac{1}{6}(x^\frac{-5}{6}-5x^\frac{-1}{6})\]
\[=\frac{1}{6}*(\frac{1}{x^\frac{5}{6}}-\frac{5}{x^\frac{1}{6}})=\frac{1}{6}*\frac{x^\frac{1}{6}-5x^\frac{5}{6}}{x^\frac{1}{6}*x^\frac{5}{6}}\]
\[=\frac{1}{6}*x^\frac{1}{6}*\frac{1-5x^\frac{4}{6}}{x}\]
\[=\frac{1}{6}*x^\frac{1}{6}*\frac{1-5x^\frac{2}{3}}{x}\]

Answer 2

so we want to find when f'=0
and when f' dne

Answer 3

f'=0 when
1-5x^\frac{2}{3}=0\]

f' dne when x=0

Answer 4

\[1-5x^\frac{2}{3}=0\]

Answer 5

\[1=5x^\frac{2}{3}\]
\[\frac{1}{5}=x^\frac{2}{3}\]

Answer 6

\[x=(\frac{1}{5})^\frac{3}{2}\]

Answer 7

\[x=\frac{1}{5^\frac{3}{2}}=\frac{1}{5\sqrt{5}}=\frac{1}{5\sqrt{5}}*\frac{\sqrt{5}}{\sqrt{5}}=\frac{\sqrt{5}}{25}\]

Answer 8

it is asking for 2 critical points

Answer 9

if you have x^2=9
this give you two solutions right?

Answer 10

-3 and 3

Answer 11

right so you you have
\[x^\frac{2}{3}=\frac{1}{5}\]
what does this give you?

Answer 12

\[x=-\frac{\sqrt{5}}{25}, \frac{\sqrt{5}}{25}\]

Answer 13

and you actually have 3 critical numbers
including x=0

Answer 14

-sqrt(5)/25

doesnt work!

Answer 15

oh wait you are right because we have even roots in our problem

Answer 16

so we only have the two
and there no more critical numbers

Answer 17

are there anymore questions?

Answer 18

oh ok! it works with x=0. Can you help me another?

Answer 19

Find the critical numbers of the function on the interval 0 ≤ θ < 2π.
g(θ) = 4 θ - tan(θ)

Answer 20

well remember we found x=0 way up there lol
a critical number is finding where f'=0 and where f' dne

Answer 21

so what do you think g' is?

Answer 22

\[(4\theta)'=?\]
\[(tanx)'=?\]

Answer 23

tanx'=sex^2(x)

Answer 24

the other one is 1

Answer 25

sorry i mean 4

Answer 26

right!
so we have
\[g'(\theta)=4-\sec^2(x)\]

Answer 27

ok then!

Answer 28

do i just input the numbers

Answer 29

now to find critical numbers we need to set g'=0 and determine where g' dne( make sure these are numbers in the domain of g)

so we have
\[g'(\theta)=4-\frac{1}{\cos^2 \theta}=\frac{4\cos^2 \theta-1}{\cos^2 \theta}\]
when is bottom 0?

Answer 30

so cosx is 0 when x=pi/2 and 3pi/2 right?

Answer 31

so those are two critical numbers
we also have to find when f'=0

Answer 32

4cos^2x-1=0
4cos^2x=1
cos^2x=1/4

Answer 33

\[cosx=\sqrt{\frac{1}{4}}=\frac{1}{\sqrt{4}}=\frac{1}{2}\]
so when does cosx=1/2 when x=?

Answer 34

do you got it?

Answer 35

60

Answer 36

and where else?

Answer 37

300 degrees or you can say 5pi/3
so we have four critical numbers

Answer 38

yes

Answer 39

2pi/3 and 4pi/3