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

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

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Grade:12

1 Answers

Jitender Singh IIT Delhi
askIITians Faculty 158 Points
9 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

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