Question

find the local minimum and maximum points and points of inflection for the function f(x) = 1/3X^3 - x^2 - 35x +17

Answer 1

\[f(x) =\frac{ 1 }{ 3 } x^3 - x^2 - 35x +17\]

Answer 2

What is the first derivative?

Answer 3

if you are allowed to use the graph, you can get the local max and min from there
https://www.desmos.com/calculator/tgu0exzdsv

Answer 4

for points of inflection we need second derivatiev f''(x).

Answer 5

F'(x) = x^2 -2x - 35

Answer 6

yes, and the second derivative?

Answer 7

F''(x) = 2x-2

Answer 8

Yes...

Answer 9

Set f''(x)=0, to solve for inflections

but there is one more little test to see if inflections are inflections

Answer 10

say you get
x=a, b
for f''(x)=0.

You need to varify that the concavity is indded changing at x=a and x=b.
let's take some number c that is sufficiently close to a (won't use epsilon or any of that), AND find find f''(x-c) and f''(x+c). If they have different signs, then the concavity indeed changed at x-a.

(Same for x=b)

Answer 11

Alright, let's go together...
f''(x)=0
2x-2=0
x=1

Answer 12

So x=1 is your (only) possible inflection point.

Answer 13

Ok. So to see if it is an actual inflection point, I am assuming we have to see if the direction of the function actually changes. How do we do this?

Answer 14

Yes, very good....

and for clarifying the direction change, we use the test I mentioned:
----------------------------
If you get
x=a for f''(x)=0.

You need to varify that the concavity is indeed changing at x=a.
let's take some number c that is sufficiently close to a (won't use epsilon or any of that), AND find find f''(x-c) and f''(x+c). If they have different signs, then the concavity indeed changed at x-a.

(Same for x=b)

Answer 15

So we can take c=0.05
f''(1-0.05) and f''(1+0.05)

if they have different signs then concaavity changes, and x=1 is indded an inflection.

Answer 16

f''(1-0.05)=f''(0.95)=2(0.95)-2
f''(1+0.05)=f''(1.05)=2(1.05)-2

Answer 17

you can tell that 2(0.95) is less than 2, so the first equation is negative. AND you can tell that 2(1.05) is greater than 2, so the second equation is positive.

Answer 18

So, fill in blank;

x=1, is indeed the __________ ____\(\bf.\)

Answer 19

if you have some questions, ask

Answer 20

X = 1 is indeed the inflection point.

Answer 21

Because there is a sign change on either side of x = 1

Answer 22

That makes a lot of sense, thanks for the help.

Answer 23

Yw