# Thread: Applications of Differentiation - Several Problems

1. ## Applications of Differentiation - Several Problems

Just need help on these problems.

1.
Suppose that 3 < than or equal to (don't know how to do the sign) $f'(x)$ < than or equal to 5 for all values of x. Show that 18 < than or equal to [tex] f(8) - f(2) < than or equal to 30.

For all of the below:
(a) Find the intervals of increase or decrease.
(b) Find the local maximum and minimum values.
(c) Find the intervals of concavity and the inflection points.

2. $h(x) = x^5 - 2x^3 + x$

3. $f(t) = t + cos(t)$, -2pi < or equal to t < or equal to 2pi

2. These are wonderful problems for teaching concepts.
What have you done to solve them?

3. Originally Posted by Plato
These are wonderful problems for teaching concepts.
What have you done to solve them?
I don't really know how to solve them. Well I'll put my work for the second one (which is probably all wrong) for the last two but the first one I do not understand at all.

2. $h(x) = x^5 - 2x^3 + x$

I took the derivative and got

$h'(x) = 5x^4 - 6x^2 + 1$

I then factored the derivative out and got

$0 = (5x^2 - 1)(x^2 - 1)$

$x = +/- \sqrt\frac{1}{5}; +/- 1$ - I know that these are the critical numbers

Then I took the second derivative which is

$h''(x) = 20x^3 - 12x$ - This has something to do with it's concavity

That's about as far as I got for this problem and I did the same thing with the other problem. I do not understand how I find the interval of decreasing, increasing or the interval of the Concavities.

4. I don't really need help on the third one. If someone can help with the second I may be able to do the third on my own. I still don't know what to do with the first one though.

5. The sign (positive or negative) of the first derivative tells you whether the original function is increasing (positive) or decreasing (negative). Where you have a critical value, the first derivative is 0. If on either side of this 0, the sign of the first derivative changes, you have either a relative minimum or a relative maximum. If it goes from negative to positive, you have a relative minimum. If it goes from positive to negative, you have a relative maximum. You can picture it like this:

If it goes from positive to negative, it is going from increasing to decreasing like this / => \ (Crude, I know.)

If it goes from negative to positive, it is going from decreasing to increasing like this \ => /

For your problem, the critical values of $h'(x) = 5x^4 - 6x^2 + 1$ are $+/- \sqrt\frac{1}{5}\; +/- 1$

Test on either side of all 4 critical values.

On either side of -1 and $\sqrt\frac{1}{5}\$, h'(x) goes from positive to negative, therefore you have a relative maximum.

On either side of 1 and $-\sqrt\frac{1}{5}\$, h'(x) goes from negative to positive and therefore has a relative minimum.

Those are the answers to part b).

6. Still kind of confusing to me, but two are relative mins and two are relative maxes?

7. Well B is sort of easier (still have some problems understanding it), but I still have no idea how to find either A and C. I don't know how to get the answers to either of them. Any help? also no one has helped me with the first problem.

Edit: Figured out how to do A. I should just make a chart for this so I won't get confused. All I need help with is the first problem and how to do C.

8. Find concavity:

1. Find the second derivative and set it equal to zero. This will give you your inflection points.

2. Using those inflection points, find your intervals. For example, if a second derivative gave the inflection points 0 and 2, the intervals would be: -infinity < x < 0, 0 < x < 2, and 2 < x < infinity.

3. Choose a point for each of the intervals. Plug it into the second derivative. Greater than zero means it concaves upward, less than zero = downward.

I don't know if that was helpful or not, but that's pretty much it step-by-step.

Find concavity:

1. Find the second derivative and set it equal to zero. This will give you your inflection points.

2. Using those inflection points, find your intervals. For example, if a second derivative gave the inflection points 0 and 2, the intervals would be: -infinity < x < 0, 0 < x < 2, and 2 < x < infinity.

3. Choose a point for each of the intervals. Plug it into the second derivative. Greater than zero means it concaves upward, less than zero = downward.

I don't know if that was helpful or not, but that's pretty much it step-by-step.
Seems pretty straightforward. Thanks

All I need now is help with the first problem and I'm done.

10. I didn't want to make a new topic for this but I still need some assistance with the first problem:

1. Suppose that 3 < than or equal to (don't know how to do the sign) $f'(x)$ < than or equal to 5 for all values of x. Show that 18 < than or equal to [tex] f(8) - f(2) < than or equal to 30.

11. Assume that ${f'}$ is an integrable function $\left[ {2,8} \right]$.
Then $\begin{array}{l}
\int\limits_2^8 {3dx} \le \int\limits_2^8 {f'(x)dx} \le \int\limits_2^8 {5dx} \\
\left. {3x} \right|_2^8 \le f(8) - f(2) \le \left. {5x} \right|_2^8 \\
\end{array}$

12. We haven't used that notation in class before so I'm kind of confused.

13. Originally Posted by FalconPUNCH!
We haven't used that notation in class before so I'm kind of confused.
This is an example of the fact that you should always tell us what you have to use.
In this case it must be the mean value theorem.
$\left( {\exists c \in \left[ {2,8} \right]} \right)\left[ {f'(c) = \frac{{f(8) - f(2)}}{{8 - 2}}} \right] \Rightarrow \quad 3 \le \frac{{f(8) - f(2)}}{{8 - 2}} \le 5
$
.

Now you finish it off.