Question

need help finding critical pts of a function

Answer 1

\[f(x)=x^\left( 2/3 \right) (4-x^2)\]

Answer 2

tale the derivative, get the critical points where the derivative is undefined and also where it is 0

Answer 3

I know to take the derivative of the function
\[ f \prime (x)=\frac{2}{3}x^\left( -1/3\right)(4-x^2)-2x^\left( 5/3 \right)\]

Answer 4

ick

Answer 5

Using my calculator I found the zeros at -1 and 1

Answer 6

rewrite as a fraction without exponential form and you will see it better

Answer 7

we can do the algebra if you like

Answer 8

ok

Answer 9

also dont forget to test the endpoints.. or was that just for max/min?

Answer 10

\[f(x)=x^{\frac{2}{3}}(4-x^2)\] dont use product rule, multiply out and get
\[f(x)=4x^{\frac{2}{3}-x^{\frac{5}{3}}\]

Answer 11

\[f \prime (x)= \frac{2(4-x^2)}{3x^\left( 1/3 \right)}-2x^\left( 2/3 \right)\]

Answer 12

\[f(x)=4x^{\frac{2}{3}}-x^{\frac{5}{3}}\]

Answer 13

ok I see

Answer 14

solving for the zeros

Answer 15

I need to take the derivative using power rule

Answer 16

still not a nice looking problem

Answer 17

no but then we write it not in exponential form

Answer 18

goddamn i am not getting latex preview so keep making typos

Answer 19

x=1.6

Answer 20

x=1.6 or 8/5

Answer 21

now subtract by multplying top and bottom of second fraction by
\[\sqrt[3]{x}\]

Answer 22

still ugly how am I to find the solutions without a calculator

Answer 23

we do a tiny bit more algebra

Answer 24

was that step to get a common denominator?

Answer 25

btw i made a mistake, it should be
\[f'(x)=\frac{8}{3\sqrt[3]{x}}-\frac{8\sqrt[3]{x^2}}{3}\]

Answer 26

yes we get a common denominator by multiplying second term top and bottom by
\[\sqrt[3]{x}\]

Answer 27

the cube root goes and you get
\[\frac{8-8x^2}{3\sqrt[3]{x}}\]

Answer 28

are you sure you made a mistake? I think it is 5/3

Answer 29

\[\frac{8(1-x^2)}{3\sqrt[3]{x}}\]

Answer 30

now it is clear that the zeros are at x = -1 and x = 1

Answer 31

no i didn't add the exponents correctly
\[\frac{2}{3}+2=\frac{8}{3}\]

Answer 32

\[f(x)=4x^{\frac{2}{3}}-x^{\frac{8}{3}}\]

Answer 33

\[f'(x)=\frac{8}{3}x^{-\frac{1}{3}}-\frac{8}{3}x^{\frac{5}{3}}\]

Answer 34

\[f'(x)=\frac{8}{3\sqrt[3]{x}}-\frac{8\sqrt[3]{x^5}}{3}\]

Answer 35

\[\frac{8-8x^2}{3\sqrt[3]{x}}\]

Answer 36

when we take the derivative then it is
\[f \prime (x) = \frac{8}{3x^\left( 1/3 \right)}-\frac{8x^\left( 5/3 \right)}{3}\]

Answer 37

right you are

Answer 38

multiply top and bottom of second one by
\[x^{\frac{1}{3}}\] and you will get it

Answer 39

ok I see it but we are solving for the solution
don't I just take the numerator and set it equal to zero

Answer 40

8-8x^2=0

Answer 41

yes, once you have it written as one fraction

Answer 42

exactly
\[8-8x^2=0\]
\[1-x^2=0\]
\[(1-x)(1+x)=0\]
\[x=-1, x=1\]

Answer 43

so critical points are x = -1, x = 1, and also x = 0 because the derivative is undefined there

Answer 44

I was trying to determine max and min

Answer 45

on what interval?