Originally Posted by

**mark** hi i've got a question and i'm coming up with a slightly different answer to my textbook so could someone please show me what i'm doing wrong:

find the set of values of x for which the function h, where $\displaystyle h(x) = 2 - (x + 3)^2$ is increasing

well the first thing i did was expand the brackets and differentiate:

$\displaystyle h(x) = 2 - x^2 + 6x + 9$ then $\displaystyle \frac{dh}{dx} = -2x + 6$

then arranged it like so: $\displaystyle -2x + 6 > 0 \implies -2x > -6 \implies x > -3$

when you divide both sides of an inequality by a negative value, the direction of the inequality changes.

the books answer is $\displaystyle x < -3$

any help would be appreciated, thanks