1. ## increasing function

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: $\displaystyle -2x + 6 > 0 \implies -2x > -6 \implies x > -3$

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

any help would be appreciated, thanks

2. 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
...

3. oh i see, i thought the minus 2 would turn to a plus 2 when it was moved to the other side of the inequality? does it just stay the same then when you're using it to divide?

4. Originally Posted by mark
oh i see, i thought the minus 2 would turn to a plus 2 when it was moved to the other side of the inequality? does it just stay the same then when you're using it to divide?
your algebra skills need some practice ...

after looking at your problem again, I see another mistake.

$\displaystyle h(x) = 2 - (x+3)^2$

$\displaystyle h(x) = 2 - (x^2 + 6x + 9)$

$\displaystyle h(x) = 2 - x^2 - 6x - 9$

$\displaystyle h'(x) = -2x - 6$

$\displaystyle -2x - 6 > 0$

$\displaystyle -2x > 6$

$\displaystyle \frac{-2x}{-2} > \frac{6}{-2}$

$\displaystyle x < -3$

also, note that you did not need to expand the (x+3)^2 in the first place ...

$\displaystyle h(x) = 2 - (x+3)^2$

$\displaystyle h'(x) = -2(x+3)$

$\displaystyle -2(x+3) > 0$

$\displaystyle x+3 < 0$

$\displaystyle x < -3$

x < -3

5. yeah it does need some practice. cheers