How do you find the domain of a function?

Question

It's been years since I studied algebra, and it all seemed to have leaked out of my head. Now I'm studying Power Functions and Function Operations, and I'm stuck on finding the domain of these functions. I just don't remember anything about finding domain and range. Here's an example:

Let f(x) = x² - 5x + 8 and g(x) = x² - 4. Perform the indicated operation and state the domain.

f(x) + g(x)

= 2x² - 5x + 4 (Let me know if I got this part wrong)

And.. For the domain. Could you explain in detail how to find domain?

Answer 1

Domain of any polynomial function is the set of real numbers



Now



f(x) + g(x) =





x² - 5x + 8



plus



x²....... - 4



equals, adding like term by like term



2x² -5x + 4



so you did it correctly.



The domain of any function y=f(x) is the set of values x is allowed to equal

and the

range of a function y=f(x) is the set of values y can, and will equal when

x assumes values within the function's domain.



Simply put, the domain is the x values and the range is the y values of any

function or relation.





I wish I could give you my Upward Bound Pre Calculus handout on finding domains of

functions. It is pretty complete. It basically says that



Find the domain of f(x) is short for



"Find the LARGEST possible domain for f(x) within the REAL NUMBERS"





One basically starts by assuming the domain is the set of real numbers and then

removes all numbers that cause the function to become undefined.

Those numbers that remain form the domain.



Assuming you are not working with transcendental functions yet {trig functions and log functions}



The two most common causes for undefined are



(1) ZERO in denominator



(2) Negative Quantity under Square root {or other even root radical such as 4th root, 6th root, ect}





so my students are expected to remember that the domain of



f(x) = 1/x is all real numbers except x= 0 which can be written (-∞,0)U(0,∞) using

the interval notation



and that the domain of

..........__

f(x) = √x is all x greater than or equal to zero which can be written [0,∞)



and that the domain of any polynomial function is the entire set of real numbers

which can be written (-∞,∞)





Suppose one has to find the domain of





......x-2..............________

y = ————_ + √500 - 10x

.........(x+3)√x



It turns out that the domain of this is the intersection of the domains of

its component functions



Component functions are

.________

√500 - 10x



and



1/(x-3)



and



1/√(x)





The domains of these component functions are



Component functions are

.________

√500 - 10x .... 500-10x≥0 so -10x≥-500 so x≤50

which is (-∞,50]



and



1/(x-3) has domain of all real x except x=3



and



1/√(x) has domain of (0,∞)



note that that is parentheses instead of bracket because square root is

in denominator



The domain of this entire function

(I give these as extra credit on the Upward Bound PreCalc Final)



is the INTERSECTION of the domains of the component functions

and it turns out to be



(0,3)U(3, 50]



You may notice I ignored the x-2 in numerator. That is because

numerators may be zero or negative all day long, it is denominators

of zero that cause functions to become undefined

Answer 2

Power Functions And Function Operations

Answer 3

Your answer for f(x) + g(x) is correct.



In order to find the domain, we need to find out the values of x where f(x) + g(x) is defined and not defined. At this point in algebra, a function will not be defined at a value of x that causes division by zero, a negative under a radical of an even root, or a negative under a logarithm. In the case of f(x) + g(x) here, this never occurs. Therefore, the function is always defined and, hence, has a domain of all real numbers.



I hope this helps!

Answer 4

f(x) + g(x) = 2x^2 - 5x + 4

The domain is all real numbers. You can substitute any real number for x and get a real output.

The range is the tricky part.

Let h(x) = f(x) + g(x)

Since h(x) is a quadratic, h(x) has an extremum where h '(x) = 0. If h ''(x) > 0, then h(x) has a minumum. If h ''(x) < 0, then h(x) has a maximum.

h '(x) = 4x - 5

0 = 4x - 5

x = 5/4

h ''(x) = 4 > 0 for all x so we know h(x) has a minimum at x = 5/4. Therefore, h(x)'s range is

[5/4, infinity).