1. ## (EASY VIEWING) Can anyone (PLEASE) Check if my answers w/ Work are [Correct]?

2. You are correct that $\displaystyle \frac{d}{dx}\left(\frac{x + 2}{x - 3}\right) = -\frac{5}{(x - 3)^2}$.

However, to find where it is decreasing, you want where the derivative is negative.

So you are trying to solve $\displaystyle -\frac{5}{(x-3)^2} < 0$.

It should be obvious that this is always negative (except at $\displaystyle x=3$ where neither the function nor derivative are defined).

So the function is decreasing for $\displaystyle x \in \mathbf{R} \backslash \{3\}$.

3. So my answer should be "it's decreasing at x=3"? i don't get this part the way to write it. X and R with the 3
Originally Posted by Prove It
You are correct that $\displaystyle \frac{d}{dx}\left(\frac{x + 2}{x - 3}\right) = -\frac{5}{(x - 3)^2}$.

However, to find where it is decreasing, you want where the derivative is negative.

So you are trying to solve $\displaystyle -\frac{5}{(x-3)^2} < 0$.

It should be obvious that this is always negative (except at $\displaystyle x=3$ where neither the function nor derivative are defined).

So the function is decreasing for $\displaystyle x \in \mathbf{R} \backslash \{3\}$.

4. If $\displaystyle f(x) = 4e^{-x^2}$ then

$\displaystyle f'(x) = -8x\,e^{-x^2}$ and

$\displaystyle f''(x) = -8e^{-x^2} + 16x^2e^{-x^2}$.

5. Originally Posted by JohnnyBean123
So my answer should be "it's decreasing at x=3"? i don't get this part the way to write it. X and R with the 3
No, it's decreasing everywhere EXCEPT at $\displaystyle x = 3$.

$\displaystyle x \in \mathbf{R}\backslash \{3\}$ means "all real $\displaystyle x$ excluding $\displaystyle 3$.

6. Originally Posted by Prove It
If $\displaystyle f(x) = 4e^{-x^2}$ then

$\displaystyle f'(x) = -8x\,e^{-x^2}$ and

$\displaystyle f''(x) = -8e^{-x^2} + 16x^2e^{-x^2}$.

Thank you for the first problem, was #7 wrong, because the final answer was the same. I'm not sure can you explain please. Thanks.

7. You had written $\displaystyle f''(x) = -8x\,e^{-x^2} + 16x^2e^{-x^2}$, not $\displaystyle f''(x) = -8e^{-x^2} + 16x^2e^{-x^2}$...

8. Originally Posted by Prove It
You had written $\displaystyle f''(x) = -8x\,e^{-x^2} + 16x^2e^{-x^2}$, not $\displaystyle f''(x) = -8e^{-x^2} + 16x^2e^{-x^2}$...
oh i appreciate it, you dropped the X because of making 8 Xe ^(-x)^^-2 the derivative?

9. When using the product rule, $\displaystyle u = -8x \implies u' = -8$ and $\displaystyle v = e^{-x^2} \implies v' = -2x\,e^{-x^2}$.

So $\displaystyle u'v + uv' = -8e^{-x^2} + 16x^2e^{-x^2}$.

10. oh i see it now, thanks. Can you please check # 3, 8, 13, please. I'm not too sure about those. If you can, don't want to take too much of your time. I appreciate the help you've already given me.

Originally Posted by Prove It
When using the product rule, $\displaystyle u = -8x \implies u' = -8$ and $\displaystyle v = e^{-x^2} \implies v' = -2x\,e^{-x^2}$.

So $\displaystyle u'v + uv' = -8e^{-x^2} + 16x^2e^{-x^2}$.

11. #3 is correct.

#8 is incorrect. If $\displaystyle f'(x) = \frac{1}{(x+1)^2} = (x+1)^{-2}$ then $\displaystyle f''(x) = -2(x+1)^{-3} = -\frac{2}{(x+1)^3}$.

So $\displaystyle f''(2) = -\frac{2}{(2+1)^3} = -\frac{2}{27}$.

12. Originally Posted by JohnnyBean123
Too many questions in one thread. Please post the ones you are still stuick on in new threads (after reading the appropriate forum rule). It would also be easier for several reasons if you typed the questions and your attempt in answering them.