1

For the function $f(x)=x^3-9x^2$, how do I determine concavity?

I know that for $f'(x) < 0$ the function is concave up when $f''(x)>0$, and concave down when $f''(x) <0$, etc...

$f'(x) = 3x^2-18x$,

$f''(x) = 6x-18$.

Do I need to evaluate at a specific point like zero to determine sign?

$f'(x) > 0$ when $x > 6$,

$f''(x) > 0$ when $x > 3$.

So on a number line for $x$ values, this equation is concave down for values to the left of 3 and concave up for values to the right and including 3?

J. W. Perry
  • 5,447
  • 3
  • 22
  • 28
user2849499
  • 51

1 Answers1

1

There are 2 points at which $f^{\prime}(x)=0$. They are $x=0,x=6$. You need to see the second derivatives at these points and since these are the only zeros of the function you can determine the concavity by viewing the second derivatives there. When $f^{\prime \prime}(x)$ changes its sign from negative to positive, concavity shifts the other way and that has already been found out by you as $x=3$. So essentially the function is Concave up (since at $x=0$, the second derivative is negative) in the interval $(-\infty,3]$ and concave down in the interval $[3,\infty)$.

Sudarsan
  • 876