Question

Use the first derivative to find all critical points and use the second derivative to find all inflection points. Use a graph to identify each critical point as a local maximum, a local minimum, or neither.
f(x)=x^5-25x^4+50. Enter answers in increasing order.

Answer 1

you there?

Answer 2

Derivative: f'(x) = 5x^4-100x^3

Answer 3

f'(x) = 5x^4 - 100x^3 = 0 at critical points
5x^3(x - 20) = 0
x = 0 and x = 20 are critical points

Answer 4

Critical points are the points where (1) the derivative is zero or (2) the derivative doesn't exist.

Answer 5

and?

Answer 6

i also need inflection point and whether the criticals are local maximum or minimum

Answer 7

The second derivative is:
20x^3-300x^2

Answer 8

20x^3-300x^2=0
x^2(20x-300)=0
So, the zeroes of the second derivative are: x=0 and x=300/20=15

Answer 9

second derivative is 20x^3 - 300 x^2
x=0 gives value of 0

x = 20 gives positive value so this is a minimum

Answer 10

still not aswring the full question

Answer 11

at x = 0 we could have point of inflection or a maximum

Answer 12

ok so could u please just put every answer together at one time?

Answer 13

check f'(x) around region of x = 0

f'(-0.1) = 5(-0.1)^4 - 100(0.01)^3 is positive
f'(0.01) is negative

thus x = 0 gives a local maxm

Answer 14

ok you have a local maximum at (0,50) and a local minimum at (20, -799950)

Answer 15

thank u! but what about the inflection point. What is that?

Answer 16

i'm not sure if there is one - let me check

Answer 17

thnks

Answer 18

no - if second derivative = 0 as it is at x = 0 it could be a point of inflection but the slopes around x = 0 show a maximum - i plotted the function using wolfram alpha which confirmed this

there is only 1 maxm and 1 minimum

Answer 19

so no?

Answer 20

You can verify that x = 0 is not an inflection point by checking the values of the second derivative around x = 0
20(0.01)^3 - 300 (0.01)^2 = -0.02998
20(-0.01)^3 - 300 (-0.01)^2 = -0.03002
Since around x = 0 the second derivative does not change sign (it is always negative) it is not an inflection point.

Answer 21

yes

Answer 22

ok, no inflection point

Answer 23

We still have to check whether there is an inflection point at x =15.

Answer 24

ok

Answer 25

u mean 20

Answer 26

I mean 15, because it is the other zero of the second derivative. Let's check the signs of the second derivative around x = 15.
20(14.998)^3 - 300(14.998)^2 = -8.99760016000073
20(15.002)^3 - 300(15.002)^2 = 9.002400159995886
It changes signs around x=15, therefore it is an inflection point.

Answer 27

so x=15 is the inflection point?

Answer 28

(15, -506200) is the inflection point.

Answer 29

which one is it? x=?

Answer 30

15?

Answer 31

Correct.

Answer 32

im going to go try all this and giv u feedback

Answer 33

You can confirm that with Wolfram|Alpha:
http://www.wolframalpha.com/input/?i=inflection+points+of+x^5-25x^4%2B50

Answer 34

k one sec thanks

Answer 35

the critical points are 0 with local max and 20 with local min?

Answer 36

Correct.

Answer 37

ok, thank you!

Answer 38

You're welcome.

Answer 39

than you also anonymous - i've learned something new about points of inflection today

Answer 40

You're welcome, joeywhite.