Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Please check your email for login details.
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

Image is attached . How to draw graph of this function?

Image is attached . 
How to draw graph of this function?

Question Image
Grade:12

1 Answers

Jitender Singh IIT Delhi
askIITians Faculty 158 Points
6 years ago
Ans:
f(x) = x^{3}+x^{2}+3x+sin(x)
Domain is given as R.
f(0) = 0
function passes through origin.
f'(x) = 3x^{2}+2x+3+cos(x)
f'_{1}(x) = 3x^{2}+2x+3
f'_{1}(x) \geq \frac{8}{3}
\Rightarrow f'(x) > 0
So, function is always increasing.
f''(x) = 6x+2-sin(x)
From here, find the region where f’’(x) is +ve & -ve.
f''(x) >0
-> Concave Up
f''(x) < 0
-> Concave down
Thanks & Regards
Jitender Singh
IIT Delhi
askIITians Faculty

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

ASK QUESTION

Get your questions answered by the expert for free