1. ## incresing/decreasing functions

show that the function $f(x)=\frac{-x^3}{3} + x^2 -x +14$ is a decreasing function

I just cant prove that $f'(x) < 0$

2. Hello,
Originally Posted by Musab
I just cant prove that $f'(x) < 0$
In this case $f'$ is a quadratic function. If you manage to factor it it'll be easy to show that $f'(x)<0$.

3. yes,

but I still cant solve it

4. Why, what have you got?

5. ## Decreasing function

It's easy bro. Once you've gotten $f'(x)$, just put it $= 0$ to get critical points. Then show that $f'(x) < 0$ on either side of the critical point.

I hope that helps.

me07.

6. Originally Posted by Musab
show that the function $f(x)=\frac{-x^3}{3} + x^2 -x +14$ is a decreasing function

I just cant prove that $f'(x) < 0$
function $f(x)=\frac{-x^3}{3} + x^2 -x +14$

differentiate the function $f'(x)=-x^2+2x-1$

$f'(x)=-(x^2-2x+1)$

$f'(x)=-(x-1)^2$

$f'(x)=-(x-1)^2<0;\;always$

since $(x-1)^2\geq 0, \text {being a perfect square}$

So, f(x) is a decreasing function.

7. Originally Posted by Shyam
function $f(x)=\frac{-x^3}{3} + x^2 -x +14$

differentiate the function $f'(x)=-x^2+2x-1$

$f'(x)=-(x^2-2x+1)$

$f'(x)=-(x-1)^2$

$f'(x)=-(x-1)^2<0;\;always$

since $(x-1)^2\geq 0, \text {being a perfect square}$

So, f(x) is a decreasing function.
$f'(x)=-(x-1)^2<0;\;always$

Just wondering I don't think you can say always because when x = 1 this happens

$0 < 0$

which is false

so it should be

$f'(x)=-(x-1)^2<0;\; \text {always, except when x = 1}$

Also at

$f'(1)= 0$

so this is niether decreasing or increasing.

8. A *strictly* decreasing function has its derivative less than zero everywhere.

A function which is just "decreasing" has its derivative less than or equal to zero everywhere.

That is, its gradient is allowed to be zero at certain points, just never positive.

You've been asked to show that this function is decreasing, not that it is strictly decreasing.

It's a subtle point but one worth taking on board.