Erfana Beevee Binte Hassan Shariff's answer to Madeline's Secondary 4 A Maths Singapore question.
done
{{ upvoteCount }} Upvotes
clear
{{ downvoteCount * -1 }} Downvotes
K=-8 is rejected because when k=-8, y= x(x-8)^3. When k=-8, dy/dx is 0 when x=8 and not when x=2. Dy/dx=o when k=-2 when x=2. Hence, k=-2 and k=-8 is rejected.
Hope this helps.
Hope this helps.
Date Posted:
1 month ago
My dy/dx is (x+k)^2.(4x+k).
When i sub in x=2, k=-2 and x=2, k=-8, both get 0
When i sub in x=2, k=-2 and x=2, k=-8, both get 0
Pls take a look at my working/explanation. Hope it helps. There is slightly more to do to determine the value of k in this question. For an intermediate question like this, more conditions could be placed in the question so that it is less challenging than it is supposed to be. Anyway, you have gotten the concepts of maxima and minima right, so no worries :)
360 Tutoring Program