Asked by purple1811
CURVE SKETCHING
π¦ = βπ₯3\3β 2π₯2 + 5π₯ β 2
ANSWER: πΆπ: π₯ = 1,β5 , IP : π₯ = β2, Decreasing on (ββ,β5) & (1, β), Increasing on (β5,1) , CU on
(ββ,β2) CD on (β2, β), Minimum point (β5,β106\3) , Maximum point (1,2\3)
I WANT THE STEPS PLEASE
π¦ = βπ₯3\3β 2π₯2 + 5π₯ β 2
ANSWER: πΆπ: π₯ = 1,β5 , IP : π₯ = β2, Decreasing on (ββ,β5) & (1, β), Increasing on (β5,1) , CU on
(ββ,β2) CD on (β2, β), Minimum point (β5,β106\3) , Maximum point (1,2\3)
I WANT THE STEPS PLEASE
Answers
Answered by
oobleck
y' = -x^2 - 4x + 5 = -(y+5)(y-1)
y'=0 at x = -5,1
y'<0 for x < -5 or x>1; y'>0 for -5<x<1
y" = -2x-4
y"(-5) > 0, so y is concave up (minimum)
So now you have the steps for these three. Go back and review the section where it covers 1st and2nd derivative tests.
y'=0 at x = -5,1
y'<0 for x < -5 or x>1; y'>0 for -5<x<1
y" = -2x-4
y"(-5) > 0, so y is concave up (minimum)
So now you have the steps for these three. Go back and review the section where it covers 1st and2nd derivative tests.
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.