Thread: Finding the intervals . Urgent.

1. Finding the intervals . Urgent.

Determine the intervals where the function is increasing and where it is decreasing. (Select all that apply.)

Increasing: (e,) (-,e) (-,0) ∪ (e,) (0,) (-,) (-,0) (0,e)
Decreasing: (0,) (-,e) (0,e) (-,0) (-,0) ∪ (e,) (e,) (-,)

Determine the interval(s) where the function f(x) = x2e-x is increasing and where it is decreasing.

Increasing on: (Select all that apply.)
(-, 2) nowhere (-, ) (0, 2) (2, ) (0, )

Decreasing on: (Select all that apply.)

(-, 2) (-, 0) (-, ) nowhere (0, 2) (2, )

2. I would recommend that you start by graphing the function of a Ti-83 or higher. If that does not help please post again and I'll try to help more.

3. i dont have a graphing calc and this is due in 1 hr

4. Shazzam!!

Cool math .com - Online Graphing Calculator - Graph It!

A graphing calculator for you.

6. Dont you have to take the derivative of it first?

7. Answer these questions while looking at the graph for each one:

- Do I have a good picture of this function?
- Where is this function defined?
- Where are the maxima of this function?
- Where is the minima of this function?
- What is the end behavior
- On the left?
- On the right?

8. Just help me i havehalf hr gotta go sleep

9. I have helped you. I have a great deal of work as well, however it is against academic honesty to tell you the answers outright. I strongly recommended that you purchase a graphing calculator asap or discuss this with your teacher.

However, as you do not have a calculator you can use the point at which the derivative of these functions change signs to determine a maximum or minimum according to how the signs change (+ to - versus - to +). You can model the end behavior of logarithmic functions by finding the limits as x approaches zero and x approaches infinity. For the other problem I recommend finding the limits as x approaches positive or negative infinity.

10. There were 28 other problems tgat i solved myself but i need help on these cuz i am lost and dont have enuff time.

11. The second prob function is x^2 e^-x

12. Hello, djdownfawl!

I must assume that you know the basics of derivatives.

. . If $\displaystyle f'(x)$ is positive, the function is increasing.
. . If $\displaystyle f'(x)$ is negative, the function is decreasing.

Determine the interval(s) where the function is increasing and where it is decreasing.

. . $\displaystyle f(x) \:=\:\dfrac{\ln x}{x}$

First, we note that $\displaystyle \,x$ must be positive.

The derivative is: .$\displaystyle f'(x) \;=\;\dfrac{x(\frac{1}{x}) - 1\cdot\ln(x)}{x^2} \;=\;\dfrac{1-\ln x}{x^2}$

Increasing: we want $\displaystyle f'(x) > 0.$

We note that the denominator, $\displaystyle \,x^2$, is always positive.

So we want: .$\displaystyle 1 - \ln x \:>\:0 \quad\Rightarrow\quad -\ln x \:>\:-1$

Multiply by $\displaystyle \text{-}1\!:\;\;\ln x \:<\:1 \quad\Rightarrow\quad x \:<\:e$

Therefore: .$\displaystyle \begin{Bmatrix}\text{Increasing:} & (0,\,e) \\ \\[-4mm] \text{Decreasing:} & (e,\,\infty) \end{Bmatrix}$

Determine the interval(s) where the function is increasing and where it is decreasing.

. . $\displaystyle f(x) \:=\:x^2e^{-x}$

The derivative is: .$\displaystyle f'(x) \;=\;x^2(\text{-}e^{-x}) + 2xe^{-x} \;=\;\dfrac{x(2-x}{e^x}$

The denominator is always positive.

The numerator is positive when the two factors are both positive or both negative.

. . Both positive: .$\displaystyle x > 0\,\text{ and }\,2-x\,>\,0$
. . . . . . . .Then: .$\displaystyle x > 0\,\text{ and }\,x \,<\,2 \quad\Rightarrow\quad (0,\,2)$

. . Both negative: .$\displaystyle x < 0\,\text{ and }\,2-x\,<\,0$
. . . . . . . . Then: .$\displaystyle x < 0\,\text{ and }\,x\,>\,2$ . . . impossible.

Therefore: .$\displaystyle \begin{Bmatrix}\text{Increasing:} & (0,\,2) \\ \text{Decreasing:} & (-\infty,\,0) \:\cup\:(2,\,\infty) \end{Bmatrix}$

13. Thank you for your reply and your time sir. But it was too late and my assignment got locked.

14. Originally Posted by djdownfawl
Thank you for your reply and your time sir. But it was too late and my assignment got locked.
Teachers expect questions that form part of an assessment that contributes towards the final grade of a student to be the work of that student and not the work of others.

Originally Posted by Soroban
Hello, djdownfawl!

I must assume that you know the basics of derivatives.

. . If $\displaystyle f'(x)$ is positive, the function is increasing.
. . If $\displaystyle f'(x)$ is negative, the function is decreasing.

First, we note that $\displaystyle \,x$ must be positive.

The derivative is: .$\displaystyle f'(x) \;=\;\dfrac{x(\frac{1}{x}) - 1\cdot\ln(x)}{x^2} \;=\;\dfrac{1-\ln x}{x^2}$

Increasing: we want $\displaystyle f'(x) > 0.$

We note that the denominator, $\displaystyle \,x^2$, is always positive.

So we want: .$\displaystyle 1 - \ln x \:>\:0 \quad\Rightarrow\quad -\ln x \:>\:-1$

Multiply by $\displaystyle \text{-}1\!:\;\;\ln x \:<\:1 \quad\Rightarrow\quad x \:<\:e$

Therefore: .$\displaystyle \begin{Bmatrix}\text{Increasing:} & (0,\,e) \\ \\[-4mm] \text{Decreasing:} & (e,\,\infty) \end{Bmatrix}$

The derivative is: .$\displaystyle f'(x) \;=\;x^2(\text{-}e^{-x}) + 2xe^{-x} \;=\;\dfrac{x(2-x}{e^x}$

The denominator is always positive.

The numerator is positive when the two factors are both positive or both negative.

. . Both positive: .$\displaystyle x > 0\,\text{ and }\,2-x\,>\,0$
. . . . . . . .Then: .$\displaystyle x > 0\,\text{ and }\,x \,<\,2 \quad\Rightarrow\quad (0,\,2)$

. . Both negative: .$\displaystyle x < 0\,\text{ and }\,2-x\,<\,0$
. . . . . . . . Then: .$\displaystyle x < 0\,\text{ and }\,x\,>\,2$ . . . impossible.

Therefore: .$\displaystyle \begin{Bmatrix}\text{Increasing:} & (0,\,2) \\ \text{Decreasing:} & (-\infty,\,0) \:\cup\2,\,\infty) \end{Bmatrix}$
A shame the assignment got locked. Soroban would have got full marks.