# Derivatives/Graphing issues

• June 3rd 2010, 05:54 PM
Harryhit4
Derivatives/Graphing issues
Hi guys. I'm having trouble going through these practice problems that we've just started in Calc A. Any help would be greatly appreciated :)

The objective is to find the

a) relative maximums
b) relative minimums
c) points of inflection
d) increasing interval
e) decreasing interval
f) concave up in the interval
g) concave down in the interval
and
h)asymptotes, if they exist

This is all after finding the first and second derivatives for:

1. $3x^5 + 5x^4$
2. $2x\sqrt{16-x^2}$
and
3. $(x^2 +3)/(x-1)$

I literally could only find the derivatives for the first equation. I am having a really tough time with this and any help is greatly appreciated :) Thanks
• June 3rd 2010, 06:39 PM
tharris
Hi. First of all you need to know that to find information reguarding max/min and inc/dec - you will need the first derivative. Then to find the critical points that you will test, set the first derivative = 0 and/or find where the first derivative is undefined (this may be why you are asked about asymptotes). After finding the critical points - whenever the first derivative is positive, the function is increasing. Whenever the derivative is negative, the function is decreasing. The max occurs when the derivative changes signs from positive to negative. The min occurs when the derivative changes from negative to positive. This is because the first derivative is describing the slope of the function. For example, 1) the first derivative is: 15x^4 + 20x^3 = 0. To solve, factor: 5x^3(3x + 4) = 0 which leads you to x=0 and x=-4/3. Now check the sign (+ or -) before and after each value. A lot of people organize this on a number line or chart.
To get the points of inflection and the concavity: use the second derivative. To find points of inflections, set the second derivative = 0 or where it is undefined. Then do the same sign test. If the second derivative is positive, the function is concave up in that interval. If the second derivative is negative, the function is concave down in that interval. If there is any kind of a sign change, then that x value is where there is a point of inflection (which is where the concavity changes).
1) Second derivative: 60x^3 + 60x^2 = 0, factor: 60x^2(x+1)=0 yields x=0 and x=-1. Now check your signs.
For problem 2 - you need to use the product rule. For problem 3 - you need to use the quotient rule. (Rofl)
• June 3rd 2010, 07:05 PM
Diemo
Quote:

Originally Posted by Harryhit4
Hi guys. I'm having trouble going through these practice problems that we've just started in Calc A. Any help would be greatly appreciated :)

The objective is to find the

a) relative maximums
b) relative minimums
c) points of inflection
d) increasing interval
e) decreasing interval
f) concave up in the interval
g) concave down in the interval
and
h)asymptotes, if they exist

This is all after finding the first and second derivatives for:

1. $3x^5 + 5x^4$
2. $2x\sqrt{16-x^2}$
and
3. $(x^2 +3)/(x-1)$

I literally could only find the derivatives for the first equation. I am having a really tough time with this and any help is greatly appreciated :) Thanks

For starters, to do the derivatives you use the product rule and the quotient rule (possably the chain rule as well for the second derivatives).
Product rule:
$\frac{d}{dx}(f(x)g(x))=f(x)\frac{d}{dx}(g(x))+g(x) \frac{d}{dx}\left(f(x)\right)$ for all functions ƒ and g. In your case (No. 2) you have that $g(x)=2x$ and $f(x)=\sqrt{16-x^2}$. However, to get the derivative of $f(x)$, it will be necessary to use the chain rule.

Chain rule: If $f(x) = h(g(x))$, then
$\frac{d}{dx}f(x)= \frac{d}{d g(x)}h(g(x)) \frac{dg}{dx}$. So in this case, your $g(x)=16-x^2$, hence you have $\frac{d}{dg}g^{\frac{1}{2}}*\frac{d}{dx}g(x)=\frac {g'(x)}{2g(x)}=\frac{-2x}{2\sqrt{16-x^2}}$

For the third one, you use the quotient rule.
$\left(\frac{f(x)}{g(x)}\right)'=\frac{f'g-g'f}{g^2}$ where the prime denotes derivative with respect to x (I am getting tired of writing these things out).

Now, for what to do in the problems:
1) Local Maxima: A local maxima is a point whose first derivative is zero and whose second derivative is negative.
2) Local Minima: A local minima is a point whose first derivative is zero and whose second derivative is positive.

To find these, find the points which are extrema (either local min or max, i.e first derivative is zero) first, and then classify them (determine for each point what the value of the second derivative is at that point and from that determine whether it is a local minima or a local maxima).

3) points of inflection are where the second derivative changes sign, or where the second derivative is zero.

4) (5) An increasing (decreasing) interval is when the rate of change of your function is positive (negative). I am pretty sure, but you might want to check your notes for them two.

6 and 7)
Definition. Let f(x) be a differentiable function on an interval I.
(i)
We will say that the graph of f(x) is concave up on I if and only if f '(x) is increasing on I. (i.e f''(x) >0)
(ii)
We will say that the graph of f(x) is concave down on I if and only if f '(x) is decreasing on I. (i.e f''(x)<0)

finally, for 8) something has an asymptote if the value for that function goes to infinity at some point. In general, this happens when there is zero under the line (the denominator). Function 1 and 2 neer blow up, hence they have no asymptotes.

Hope this helps.