Asked by kavi
10.a)Solve the inequality x^3-5x^2+2x+8<0 by
i) using intervals
ii) considering all cases
I did this but my teacher said that when i solved this: x^3-5x^2+2x+8<0
(x+1)(x^2-6x+8)
(x+1)(x-4)(x-2)<0
She told me to show all the steps, but i don't know how to show more. so can you please help me to show all the works and rest of the works i did too but she said " the same answer has to happen for each method. i don't really understand that. can you help me please and thank you....
Steve explained this, but still i didn't got the steps. can someone do it for me step by step so i can understand the steps of it please.
i) using intervals
ii) considering all cases
I did this but my teacher said that when i solved this: x^3-5x^2+2x+8<0
(x+1)(x^2-6x+8)
(x+1)(x-4)(x-2)<0
She told me to show all the steps, but i don't know how to show more. so can you please help me to show all the works and rest of the works i did too but she said " the same answer has to happen for each method. i don't really understand that. can you help me please and thank you....
Steve explained this, but still i didn't got the steps. can someone do it for me step by step so i can understand the steps of it please.
Answers
Answered by
Reiny
Ok, so all the algebra work has been done
x^3-5x^2+2x+8<0
(x+1)(x-4)(x-2)<0
Quickest way:
I would look at the graph of
y = x^3-5x^2+2x+8
= (x+1)(x-4)(x-2)
I know that this is standard cubic function opening up in quadrant I and having x-intercepts of -1, 2, and 4
So a very rough sketch shows me that this curve is below the x-axis ( < 0 ) for x < - 1 or 2 < x < 4
looking at all cases , the critical values of -2,2, and 4 split the line into 4 parts :
if x < -1, try some value, e.g. x = -5
(-)(-)(+) < 0 , so yes
x < -1
if x is between -2 and 2 , e.g. x = 0
(+)(-)(-2) > 0 , so NO
if x is between 2 and 4 , e.g. x = 3
(+)(-)(+) < 0 , so YES
x is between 2 and 4
if x > 4 , e.g. x = 5
(+)(+)(+) > 0 , NO
then x < -2 OR 2 < x < 4
x^3-5x^2+2x+8<0
(x+1)(x-4)(x-2)<0
Quickest way:
I would look at the graph of
y = x^3-5x^2+2x+8
= (x+1)(x-4)(x-2)
I know that this is standard cubic function opening up in quadrant I and having x-intercepts of -1, 2, and 4
So a very rough sketch shows me that this curve is below the x-axis ( < 0 ) for x < - 1 or 2 < x < 4
looking at all cases , the critical values of -2,2, and 4 split the line into 4 parts :
if x < -1, try some value, e.g. x = -5
(-)(-)(+) < 0 , so yes
x < -1
if x is between -2 and 2 , e.g. x = 0
(+)(-)(-2) > 0 , so NO
if x is between 2 and 4 , e.g. x = 3
(+)(-)(+) < 0 , so YES
x is between 2 and 4
if x > 4 , e.g. x = 5
(+)(+)(+) > 0 , NO
then x < -2 OR 2 < x < 4
Answered by
Kaplan
Reiny help me with using intervals. u forgot to help me with that. its urgent please
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.