Having trouble with this math problem...

[tcrosson]

I have a final in this stuff tomorrow (5-7), and for the life of me I can't figure out how to work it out. This problem is off the review sheet the teacher gave us (without answers of cousre). Any help would be greatly appreciated.

f(x) =

• 3x (-2<x<1)
• 5 (x = 1)
• x^2 +1 (x>1)
  • 
    
    (a) Find the domain of the function.
    (b) Graph each function by hand.
    (c) Evaluate f(-1); f(0); f(1); f(2)
    
    Answers:
    a) x = ( , )
    C) f(-1) = ______
    f(0) = ______
    f(1) = ______
    f(2) = ______
    
    
    The graphing part I can do on my own, but the rest of it boggles me. And for the curious, it's College Algebra. Also, they're aren't any problems like this in the book.

 

[dighn]

3x (-2x<x<1)

5 (x = 1)

x^2 +1 (x>1)

is there a typo here. if not then since -2x is always < x, that condition reduces to x<1
for domain it means the range of possible x values, in this case x can be anything so x is all reals
if that is a typo, then it's -2<x<1, then x is not defined for x <= -2, so th edomain is any real larger than -2

now for evaluating the function, just check which condition applies to the x you are evaluating then plug x into the expression to the left of the condtion

for example f(-1), -1 is < 1, so obvious condition 1 is fit so you use f(x) = 3x, 3*-1 = 3 so f(-1) is -3

get it?

 

[Darien]

for a, notice where f(x) is defined. Generalize -2 < x < 1, x = 1 and x > 1 into one basic statement From these 3, what values of x are allowed?

for b, no comment -- you say you can do this

for c, note the value of x and then find the appropriate portion of the function to evaluate it with For example, the first one is f(-1). Since f(x = -1) is in -2 < x < 1, use the first equation. f(x = -1) would be -3.

 

[tcrosson]

Yes it was a typo, it's now fixed.

a) So if "x" is restricted to the domain of all three inequalities, the only possiblity for x would be (0, -1)?

c) What about for 5 (x = 1)? I assume since it's strict to (x = 1), then only (x=1), therefore the input (x) is "1" and the output (y) is "5"?

 

[jurzdevil]

edit** wrong thing

 

[dighn]

Originally posted by: tcrosson
Yes it was a typo, it's now fixed.

a) So if "x" is restricted to the domain of all three inequalities, the only possiblity for x would be (0, -1)?

c) What about for 5 (x = 1)? I assume since it's strict to (x = 1), then only (x=1), therefore the input (x) is "1" and the output (y) is "5"?

no
the condtions are
a. (-2<x<1)

b. (x = 1)

c. (x>1)

condition a defines x between -2 and 1, execlusive - ok
condtion b defines s at 1
a+b defines x between -2 exclueisve and 1 inclusive
copndtion c edfines x larger than 1
combinining all three you have x defined everywhere except smaller than -2 and at -2

so demain of x is >-2

5 (x=1) means if x is 1, f(x) is 5

 

[Savarak]

really easy...

a: x = (-2,positive infinte)
c: f(-1) is -3 because 3(-1) is = -3 when x is less than -1 and greater than -2
f(0) is 0 because 3(0) is = 0 when x is less than -1 and greater than -2
f(1) is 5 because when x=0 when x is equal to 1
f(2) is 5 because 2^2 + 1 = 5 when x is greater than 1

noticed a typo

 

[Brutuskend]

Standard math question response

Don't take offence, I can't do math and don't claim to be able to. I just think it's funny!

 

[tcrosson]

Thank you so much guys! Even you Brutuskend! With the semester's end bearing down hard, I could use a good laugh break the tension.