Find the interval on which the curve is concave up.

Question

amoodarya · Answer

find f ''
then f ''=0
then determine the sign of f ''
draw a table of f''

anonymous · Answer

$$y = \int\limits_{0}^{x} (t^3 + t^2 +1)dt$$

^^^
The Curve

amoodarya · Answer

no difference 
do that

anonymous · Answer

True because the derivative and integral can cancel each other out. Why Would I want to find the second derivative though?

anonymous · Answer

Scratch that last thought. That's how I can see if it's concave up or not. Okay thank you :)

amoodarya · Answer

$$y '=x^{3}+x^{2}+1$$

anonymous · Answer

So from the Fundamental Theorem of Calculus, I can substitute the "t" for an x.

Then I can find the derivative.
Is the first drivitve considered the second derivative after the integral is gone?

amoodarya · Answer

I am waiting to check your answer

anonymous · Answer

Okay cool thanks.

Well I've got y = x^3 + x^2 + 1
Then I need to find the second derivative

3x^2 + 2x

x = 0

When x is less than 0, It is concave up.

Wait isn't x always concave up because any value will make a positive value of the second derivative.

amoodarya · Answer

ok 
lets begin from begin 
suppose 
we do the integral
so we have$$y=f(t)=\frac{ t^{4} }{4 }+\frac{ t^{3} }{3 }+t$$

amoodarya · Answer

now apply f(x)-f(0) over y

anonymous · Answer

Which gives me x^4/4 + x^3/3 + x - 0

amoodarya · Answer

$$y=f(t)=x^{4}/{4} +x^{3}/{3}+x -0 -0 -0$$

amoodarya · Answer

solving integral give you that

amoodarya · Answer

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