Question

Can you help me solve this? http://screencast.com/t/XZ7n99PHIz

I am stuck at finding the decreasing and increasing intervals.

I cant solve the inequality that pops up

Answer 1

Did you find the critical points?

Answer 2

@.Sam. @agent0smith @ajprincess @amistre64 @bahrom7893 @ganeshie8 @LoveYou*69 @robtobey @skullpatrol

Answer 3

I just cant, the 2nd degree inequality that pops up , can't be solved.

Answer 4

@Christos what inequality? Find f', set it to zero, then find the x values for that. Check points to the left and right to see if it's increasing or decreasing.

http://www.wolframalpha.com/input/?i=derivative+x%5E4-5x%5E3%2B9x%5E2

Answer 5

You started by taking the first derivative getting
f'(x) = 4x^3 - 15x^2 + 18x

Answer 6

Then you set it equal to zero and solves for the x values. These are the points where the graph isn't increasing or decreasing.

Answer 7

I just get x = 0 as a point.

Answer 8

You can either check points to the left and right, and check the sign of f' (whether positive or negative)

or get the second derivative, plug in the x value for when f' = 0, then see if it's concave up or down to help you decide.

Note that if f'' = 0, it's an inflection point, so it's not a change in concavity.

Answer 9

@Christos Make sure you close questions that have been answered so people don't get confused about who needs help. Thanks!

Answer 10

@Christos Since you know the graph isn't changing at x = 0, check an easy value to the right of x = 0 (such as x = 1) and a value to the left of x = 0 (such as x = -1. Plug those values (x =1, -1) into f'(x). If your number is positive it is increasing at the point, if it is negative it is decreasing.

Answer 11

**I just cant, the 2nd degree inequality that pops up , can't be solved.**

\[ y' = 4x^3-15x^2+18x \]
you can look for max or mins by setting y'=0
\[ 4x^3-15x^2+18x =0 \\ x(4x^2-15x+18)=0\\ x=0 \text { or } 4x^2-15x+18=0\]

you can take the 2nd derivative y'' and see if y''>0 at x=0 (means a min) or y''<0 (a max)
the quadratic
\[ 4x^2-15x+18=0\]
has a discriminant of \( b^2 - 4 ab\) which you will find is negative. the roots are imaginary. That means the curve does not have any other min/max points

Answer 12

you can find the inflection points by finding y'' and setting y''=0

Answer 13

"Note that if f'' = 0, it's an inflection point, so it's not a change in concavity."

This should say it is a change in concavity.

Answer 14

@phi The reason it can't be solved is because the x boundaries go to positive and negative infinity.

Answer 15

there is a global min at x=0. as x increases the curve rises. You designate the interval where f is increasing like this:
(0, +inf )

on the other hand, the left side of the curve is descending toward the minimum value at x=0
the interval of the curve where f(x) is descending is (-inf, 0)

to answer parts c, d, and e first find the inflections points (this answers e)
can you do that ?