Question

Need help with inflection points.

Answer 1

Hello!

Answer 2

Okay so my question is asking to find the x-cooridnate of any relative extrema and inflection points for the function f(x)=\[9x^(1/3)+9/2x^(4/3)\]

Answer 3

the (1/3) and (4/3) are powers of the variables.

Answer 4

\[f(x) = 9x^{1/3} + \frac{9}{2}x^{4/3}\]

like this ?

Answer 5

yes

Answer 6

Okay so I've actually done some work so i want to see if so far im right

Answer 7

I found the first and second derivative for the equation but having trouble somewhere along the process

Answer 8

Okay great show me what you so far

Answer 9

\[f'(x)=3x ^{-2/3}+6x ^{1/3}\]

Answer 10

A necessary condition for relative extrema is the first derivative has to be 0

A necessary condition for inflection points is the second derivative has to be 0

so you're on right track in finding the points at which the derivatives are 0

Answer 11

\[f''(x)=-2divx ^{5/3}+2divx ^{2/3}\]

Answer 12

div is divided idk why it does it like that.

Answer 13

use ` \frac{}{} `

Answer 14

`\frac{2}{3}` gives you \(\frac{2}{3}\)

Answer 15

\[f′′(x)=−\frac{2}{x ^{5/3}}+\frac{2}{x ^{2/3}}\]

Answer 16

copy pasting has some issue

Answer 17

let me try again.

Answer 18

\[f''(x)=-\frac{2}{x^{5/3}}+\frac{2}{x ^{2/3}}\]

Answer 19

better?

Answer 20

looks good!
set the first derivative equal to 0 and solve x

Answer 21

so in all honesty i have trouble with the algebra. do you mind helping me out?

Answer 22

\[f'(x)=3x ^{-2/3}+6x ^{1/3} = 0\]

Answer 23

lets rationalize the numerator

Answer 24

\[\frac{3}{x ^{2/3}}+6x ^{1/3} = 0 \]

Answer 25

okay makes sense.

Answer 26

\[\frac{3+6x}{x ^{2/3}} = 0 \]

Answer 27

okay lol what happened to the ^1/3

Answer 28

How'd you get rid of it?

Answer 29

let me break it for you \[\frac{3}{x ^{2/3}}+6x ^{1/3} \]


left side is a fraction, and we know we need common denominator to add fraction. so lets multiply the second term by a special kind of \(\color{red}{1 ~:~\frac{x^{2/3}}{x^{2/3}}}\)

\[\frac{3}{x ^{2/3}}+6x ^{1/3}\times \color{red}{ \frac{x^{2/3}}{x^{2/3}} } \]

Answer 30

ofcourse! i understand it now.

Answer 31

\[ = \frac{3}{x ^{2/3}}+\frac{6x^{1/3}\cdot \color{red}{x^{2/3}}}{ \color{red}{x^{2/3}}} \]

Answer 32

I understand how now thanks.

Answer 33

so now we have that fraction and we solve for x?

Answer 34

yes ! it should be easy to solve x now as we can multiply by x^2/3 both sides and get rid off the denominator

Answer 35

okay so now we have. \[3+6x=x ^{2/3}\]

Answer 36

do we square root it?

Answer 37

careful, right hand side is 0
so anything multiplied by 0 is zero!

Answer 38

oh that is true! so rather than the x^2/3 we should have \[x=-\frac{1}{2}\]?

Answer 39

Looks good! x = -1/2 is your critical point

Answer 40

So now we must find the relative extrema's

Answer 41

and doing that dont we plot x=-1/2 on a line graph, get a number lower and then higher than it and plug in those numbers into the first derivative to get if the graph increases or decreases.

Answer 42

right!

Answer 43

depending on whether the number is positive or negative

Answer 44

okay one second i shall let you know.

Answer 45

So we should have a minimum at x=-1/2 correct?

Answer 46

relative minimum to be exact.

Answer 47

Perfect !

Answer 48

we could also use second derivative to figure out the same

Answer 49

But isn't that for concavity?

Answer 50

or in this case inflection point.

Answer 51

f''(-1/2) > 0 : relative minimum
f''(-1/2) < 0 : relative maximum

Answer 52

yes concavity is related to relative extrema

Answer 53

when the graph is concave down, the slopes of tangents around the relative minimum point decrease :