Linear and Quadratic functions

[kidnik6]

I have two questions on a homework assignment that I am a little confused about.

1) Given f(x)=2x-1 and g(x)=2x²-5x-3 find f/g and the domain.

I have solved f/g= 1/x+3. This is correct according to the book.
I have the domain as x cannot equal -3. This is also correct but the book also says x cannot equal 1/2. Why?

2) A rancher with 7000 yards of fencing wants to enclose a rectangular field that borders a straight highway and then wants to divide it into two plots with a fence parallel to the highway. If no fencing is needed along the highway ( fencing already exists), what is the largest area that the farmer can enclose?

I have gotten the answer 6,125,000 yd
The book says the answer is 3,062,500 yd^2. Is this the same thing given that my answer is double this number?

 

[Yogi]

Authors of books make errors from time to time.

After you draw the picture you should have 2w+2l=7000 and A=wl.

 

[kidnik6]

It is +5x. That was my mistake. How does 1/2 make the den. 0? Wouldnt the 1/2 + 3 simply make the den 3 1/2?

 

[Yogi]

If you are unsure graph it, asymptotes at x = -3 and y=0. j(1/2) = \(\displaystyle \frac{1}{3.5}\)

 

[Yogi]

The exact definition of the function is h(x) = f(x) / g(x).

So that function does not exist if g(x) = 0. With me so far? False. g(1/2) = 0 but j(1/2) = \(\displaystyle \frac {1}{3.5}\)


Furthermore, g(x) is a quadratic, which means it may equal 0 at TWO values of x.

There are several ways to find those values, the simplest one being to factor the quadratic if
a factorization is obvious. In any case, however you find them, they are 1/2 and -3.

\(\displaystyle 2(-3)^2 + 5(-3) - 3 = (2 * 9) - 15 - 3 = 18 - 18 = 0\). Don't need to plug back in and check.

\(\displaystyle 2\left(\frac{1}{2}\right)^2 + 5\left(\frac{1}{2}\right) - 3 = 2\left(\frac{1}{4}\right) + 5\left(\frac{1}{2}\right) - 3 = \dfrac{1 + 5}{2} - 3 = 3 - 3 = 0\).

Graph it if you need to. Graph blows up at -3, behaves normally at 1/2. Domain is all real numbers except -3.

 

[lookagain]

Graph it if you need to. Graph blows up at -3, behaves normally at 1/2.

> > > Domain is all real numbers except -3. < < <

No, there is no point at x = 1/2. There is a "hole" there. And it is true that x = -3 is a vertical asymptote.

The (2x - 1) factors cancel/divide out as long as x is not equal to 1/2.

The domain is (-oo, -3) U (-3, 1/2) U (1/2, oo).


Here's another example:

The graph of y = x/x has a domain of all real numbers except x = 0.

 

[Yogi]

Ah right, I guess I was thinking about limits for some reason. oops!

 

[HallsofIvy]

It might be useful here to point out that "function" is not the same as a "formula". To have a function, we do need some "rule", which we typically write as a formula, to tell us how to find y for a given value of x, and a 'domain'- the allowable values of x. Often this is the "natural domain"- the set of all values of x for which the formula can be calculated but it could as well be any subset of that. And if we change the domain, we change the function.

That is, if we talk about the "function", \(\displaystyle f(x)= x^2\), without mentioning a domain, that would, by default, be taken to be the function, using that formula, whose domain is all real numbers. But if we talk about the function \(\displaystyle g(x)= x^2\), with domain all non-negative numbers, that is a completely different function.

Similarly, \(\displaystyle f(x)= \frac{1}{x+3}\) would have natural domain "all x except -3". But the function \(\displaystyle g(x)= \frac{1}{x+3}\) with domain "all x except -3 and 1/2" is a completely different function. And it is true to say that \(\displaystyle \frac{2x- 1}{2x^2+ 5x-3}= g(x)\), NOT f(x).

I suspect that was what your text book was saying, not that "\(\displaystyle \frac{2x-1}{2x^2+ 5x- 3}= \frac{1}{x+3}\)"

 

[kidnik6]

Makes sense. I was looking at the simplified version of f/g which is 1/x+3 which in that case the domain is x cannot equal -3.

I forgot to look at how I factored it before which was (x+3)/(2x-1)(x+3). So they got the x cannot equal 1/2 from 2x-1. Thats all I needed.

Thanks for your help everyone. I was able to figure out the word problem as well.

 

[lookagain]

Makes sense. I was looking at the simplified version of f/g which is 1/x+3 ...

Don't forget your grouping symbols around the denominator. **

It should be 1/(x + 3), for example.


which in that case the domain is x cannot equal -3.

I forgot to look at how I factored it before which was (x+3)/(2x-1)(x+3).


See ** above for this also. And the expression isn't what you typed. You have the factors reversed.

It is supposed to be (2x - 1)/((x + 3)(2x - 1)).

(But the order of the factors in the denominator doesn'r matter.)

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