# Concavity

• Oct 18th 2009, 04:26 PM
Open that Hampster!
Concavity
$\displaystyle f(x) = x^q(x+6)$
$\displaystyle f'(x) = \frac{x+6}{5x^z} + x^!$
$\displaystyle f''(x) = \frac{5x^z + \frac{x+6}{4x^q}}{25x^k}$
Note: I couldn't get fractional exponents to show up properly, so I used this key: $\displaystyle q = \frac{1}{5} ; z = \frac{4}{5} ; k = \frac{8}{5}$

We need to find where f is concave up and down. There's two inflection points, one of which I've (correctly) found to be (0,0), which means that f is concave up at $\displaystyle (-\infty,0)$ and from $\displaystyle (s,\infty)$ and it's concave down from $\displaystyle (0,s)$.

But f''(x) never goes negative, and it's undefined at 0, so I'm not sure how to find s.
• Oct 18th 2009, 04:40 PM
tonio
Quote:

Originally Posted by Open that Hampster!
$\displaystyle f(x) = x^q(x+6)$
$\displaystyle f'(x) = \frac{x+6}{5x^z} + x^!$
$\displaystyle f''(x) = \frac{5x^z + \frac{x+6}{4x^q}}{25x^k}$
Note: I couldn't get fractional exponents to show up properly, so I used this key: $\displaystyle q = \frac{1}{5} ; z = \frac{4}{5} ; k = \frac{8}{5}$

We need to find where f is concave up and down. There's two inflection points, one of which I've (correctly) found to be (0,0), which means that f is concave up at $\displaystyle (-\infty,0)$ and from $\displaystyle (s,\infty)$ and it's concave down from $\displaystyle (0,s)$.

But f''(x) never goes negative, and it's undefined at 0, so I'm not sure how to find s.

Why to mess up with derivatives of first and second order of a product if you can enjoy yourself with a simple polynomial?? Write simply

$\displaystyle f(x)=x^{q+1}+6x^q$

Now evaluate the second derivative:

$\displaystyle f'(x)=(q+1)x^q + 6qx^{q-1}$

$\displaystyle f''(x)=q(q+1)x^{q-1}+6q(q-1)x^{q-2}=qx^{q-2}[(q+1)x+6(q-1)]$

Where the second derivative is positive and where negative is easier now to see, I'm sure.

Tonio