![<marquee> <marquee>](https://tex.z-dn.net/?f=%26lt%3Bmarquee%26gt%3B)
![\huge\pink{\underline{GOOD-EVENING....}} \huge\pink{\underline{GOOD-EVENING....}}](https://tex.z-dn.net/?f=%5Chuge%5Cpink%7B%5Cunderline%7BGOOD-EVENING....%7D%7D)
![\huge\bold\blue{\fbox{\fbox{\red{\mathcal{SOLVE/:THE /:ATTACHMENT......}}}}} \huge\bold\blue{\fbox{\fbox{\red{\mathcal{SOLVE/:THE /:ATTACHMENT......}}}}}](https://tex.z-dn.net/?f=%5Chuge%5Cbold%5Cblue%7B%5Cfbox%7B%5Cfbox%7B%5Cred%7B%5Cmathcal%7BSOLVE%2F%3ATHE+%2F%3AATTACHMENT......%7D%7D%7D%7D%7D)
![<body bgcolor=lime><maree><font color=red> <body bgcolor=lime><maree><font color=red>](https://tex.z-dn.net/?f=%26lt%3Bbody+bgcolor%3Dlime%26gt%3B%26lt%3Bmaree%26gt%3B%26lt%3Bfont+color%3Dred%26gt%3B)
Attachments:
![](https://hi-static.z-dn.net/files/d58/38a5417294707866210f5cb6893ea08d.jpg)
Answers
Answered by
3
• We know that,
x³/3 = 3x²/3
= x²
• Similarly,
5x²/2 = 2*5x/2
= 5x
• 6x is a constant. So, its differentiation becomes 0.
Step - I :-
Now,
y = x³/3 + 5x²/2 + 6x
⇒dy/dx = y' = x² + 5x + 0 = 0
⇒y' = x² + 5x = 0
⇒y' = x(x + 5) = 0
⇒ x = 0, or x = -5
Step - II :-
y" = dy'/dx = x² + 5x
⇒y" = 2x + 5
⇒y" = 2(0) + 5 or y" = 2(-5) + 5
⇒y" = 0 + 5 or y" = -10 + 5
⇒y" = 5 or y" = -5
5 > 0 and -5 < 0
So, 5 is minima and -5 is maxima.
Similar questions