# Concavity

• Jan 4th 2009, 01:58 PM
abclarinetuvwxyz
Concavity
The graph of f(x)= integral from 0 to x (15(t^2) - 2(t^3) + 24)dt is concave up on (a,b). Find b-a.

Okay so I know from the curve sketching lesson in AB calculus that to find concavity you take the second derivative and then set that to 0. Do I take the integral and then take the first derivative, and then the second?
• Jan 4th 2009, 02:17 PM
skeeter
• Jan 4th 2009, 03:39 PM
Soroban
Hello, abclarinetuvwxyz!

Quote:

The graph of: .$\displaystyle f(x)\;= \;\int^x_0(15t^2 - 2t^3 + 24)\,dt$ . is concave up on $\displaystyle (a,b).$]
Find $\displaystyle b-a$

The first derivative is: .$\displaystyle f'(x) \:=\:15x^2 - 2x^3 + 24$

The second derivative is: .$\displaystyle f''(x) \:=\:30x - 6x^2$

Now where is the graph concave up?
. . Where $\displaystyle f''(x)$ is positive.

So we have: .$\displaystyle 30x - 6x^2 \:>\:0\quad\Rightarrow\quad 6x(5 - x) \:>\:0$

This is true when: .$\displaystyle 0 \,<\,x\,<\,5$

Therefore, the interval is: .$\displaystyle (0,5) \quad\Rightarrow\quad b - a \:=\:5-0 \:=\:\boxed{5}$

• Jan 4th 2009, 03:49 PM
abclarinetuvwxyz
Thanks very much!