Building of Functions based on certain requirements

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June 25th 2007, 05:52 AM
lamiastella

Building of Functions based on certain requirements

Hi everyone, I'm back! :)

I did a small intro in the post I made in the Statistics section, but I guess I'll make one again. My name's Melanie and I'm a university student here in Germany, but I come from the US. Currently, we're getting ready to take our final exams, and I'm kind of lost on a few topics.

One area in general where I have a lot of problems is "Building functions based on requirements". Now, I'm not sure if this belongs in the Calculus forum, but since the course I'm taking is called Analysis (Calculus) I assume that I'm in the correct place.

So, onto my question! I must have the worst math professor there ever was - I attempted to ask for help, but she basically shot me down. Nobody else in my class understands what in the world is going on either.

So, here is the question (sorry if the wording is funny, I'm translating from German to English):

2. Wanted: The equation of a rational 3. Degree function, that has the following requirements:

http://i11.tinypic.com/52aooer.jpg

...Sorry, I don't know how to type in the math stuff here, so I just did it in word and got a screenshot.

Now, my problem is, is that I don't even know how to begin with these types of problems.

I scanned in my attempt to do the problem, but I don't know where to go once I did what I did below:

http://i19.tinypic.com/63camg9.jpg

Okay, fixed the scan.

Thank you all! You guys are really great :D
June 25th 2007, 05:53 AM
lamiastella

Okay, got a better scan! :)
June 25th 2007, 07:25 AM
lamiastella

Okay, I found my mistake. In the part on the notebook paper where it says x = 1, it's supposed to be x = -1 :(
June 25th 2007, 08:01 AM
topsquark

Okay, we've got a cubic function:
$f(x) = ax^3 + bx^2 + cx + d$
which goes through the point (1, 2), has a relative maximum at x = -1, and an inflection point $\left ( -\frac{1}{3}, -\frac{10}{3} \right )$ .

So.
(1) $f(1) = 2 = a + b + c + d$

And
$f^{\prime}(x) = 3ax^2 + 2bx + c$
so
(2) $0 = 3a - 2b + c$

And
$f^{\prime \prime} = 6ax + 2b$
so
(3) $0 = 6a \left ( -\frac{1}{3} \right ) + 2b$
and
(4) $f \left ( -\frac{1}{3} \right ) = -\frac{10}{3} = -\frac{a}{27} + \frac{b}{9} - \frac{c}{3} + d$

Condition (3) says that
$0 = 6a \left ( -\frac{1}{3} \right ) + 2b$

$0 = -2a + 2b$

$b = a$

Thus
(1) $f(1) = 2 = a + a + c + d$

(1) $2 = 2a + c + d$

(2) $0 = 3a - 2a + c$

(2) $0 = a + c$

(4) $f \left ( -\frac{1}{3} \right ) = -\frac{10}{3} = -\frac{a}{27} + \frac{a}{9} - \frac{c}{3} + d$

(4) $-\frac{10}{3} = \frac{2a}{27} - \frac{c}{3} + d$

Equation (2) says:
$0 = a + c$

So
$c = -a$

Thus
(1) $2 = 2a + (-a) + d$

(1) $2 = a + d$

(4) $-\frac{10}{3} = \frac{2a}{27} - \frac{-a}{3} + d$

(4) $-\frac{10}{3} = \frac{11a}{27} + d$

Now, equation (1) says:
$2 = a + d$

$d = 2 - a$

So
$-\frac{10}{3} = \frac{11a}{27} + (2 - a)$

$-\frac{16}{3} = -\frac{16a}{27}$

$a = 9$

$b = 9$

$c = -9$

$d = -7$

So
$f(x) = 9x^3 + 9x^2 - 9x - 7$
has all the desired properties.

-Dan
June 25th 2007, 08:14 AM
lamiastella

Dan, I cannot thank you enough!!

Thank you for taking the time to explain it to me like you did. It really helps me when people explain it slowly and with a lot of detail. Again, I really appreciate it!

Now that I have seen how it's done, I can go work on the other problems that we were given. If I do it slowly, then I should be fine!