Question

Help! Trying to solve this equation without graphing - how can I do that? f(x)= x^2-1 {x: -1 < x < 5} Then the range of f(x) is?

Answer 1

Are you trying to find the roots of f(x) = x^2 - 1 ?

Answer 2

If so, then solve x^2 - 1 = 0.
You can try factoring first.

Answer 3

No, I'm trying to find the range of the function f(x)

Answer 4

The domain is written {x: -1 < x < 5} so given that and the function itself f(x) = x^2 -1, how do I find the range?

Answer 5

The solution tells me this - which I don't exactly understand:
. f(x) is not monotonic, increasing in the interval (–1, 5), so we cannot simply use (–1, 5) to get the range. The monotonic interval is (0, 5); therefore the range is (–1, 24). In this case a graph is very useful to find out the monotonic interval. The graph of the quadratic function is as following:

I obviously get that the answer is (-1, 24) but I'm not sure I understood the thought process.

Answer 6

You wrote "Trying to solve this equation without graphing - how can I do that?"
I thought you were trying to solve the equation because you said so.

Answer 7

Sorry, I misspoke. I just need to find the range.

Answer 8

Ok. No problem.
f(x) = x^2 - 1 is similar to f(x) = x^2, but it is shifted 1 unit down from f(x) = x^2

Answer 9

f(x) = x^2 is a parabola that opens up with vertex (0, 0). As x increases from x = 0, f(x) = x^2, so f(x) is increasing as the square of x.
For x < 0, you just need to think of the mirror image of the side with x > 0.
f(x) = x^2 is symmetric with respect to the y-axis, since you can replace x with -x whitout changing the equation.
Now think of your specific function, f(x) = x^2 - 1.
It is just the same as f(x) = x^2 shifted one unit down.
It is still symmetric wrt the y-axis.
The vertex now is (0, -1)

Answer 10

I think I understand how -1 is one end of the range, but how did we get 24?

Answer 11

Could use some help if you can @mathstudent55

Answer 12

At the high end of the domain, x = 5.
Then f(x) = x^2 - 1
f(5) = 5^2 - 1
f(5) = 25 - 1
f(5) = 24

Answer 13

Ohhh! that makes a lot of sense.

Answer 14

this is the way I look at it:
\[-1<x<5 \\ \text{ squaring } x \text{ means } 0<x^2<25 \\ \\ \text{ then subtracting 1 on add sides of the inequality } \\ 0-1<x^2-1<25-1 \\ -1<x^2-1<24 \\ -1<f(x)<24 \\ -1<y<24\]

Answer 15

all not add*

Answer 16

How does -1 squared equal 0?

Answer 17

Wouldn't it = 1?

Answer 18

it doesn't but x^2>=0 for all x

Answer 19

oops I should have put an equal sign

Answer 20

\[0 \le x^2 <25 \\ 0-1 \le x^2-1 <25 -1 \\ -1 \le x^2-1 <24 \\ \]

Answer 21

Okay this does make sense, except for the -1 squared part. Still not sure why -1*-1 would = 0?

Answer 22

i said (-1)^2 is not 0

Answer 23

x^2>=0 for all x

Answer 24

sorry, I'm kind of slow sometimes haha. how do we know that?

Answer 25

how do we know a real number square will give us a positive number or 0?

Answer 26

like is that just a rule that x is always greater than or equal to 0?

Answer 27

no I never said x is greater than 0

Answer 28

oh I see what you mean - any number squared is greater than 0

Answer 29

or 0

Answer 30

0^2=0

Answer 31

(-1)^2=1 >0

Answer 32

okay cool... I think this makes sense now. Thank you for your help!

Answer 33

i have to go

Answer 34

If anyone else sees this, I am still not sure about how this would work in another situation - like, can I take a similar situation and assume -1<x<5 = 0<x^2<25?

Answer 35

I just don't want to skip past this without understanding it completely.

Answer 36

let's look at something...
\[-1<x<5 \text{ is the same as } -1<x<0 \cup x=0 \cup 0<x<5 \\ \\ \\ \text{ so let's look at this bit by bit } \\ -1<x<0 \implies 1>-x>0 \text{ notice I change the direction of the inequality } \\ \text{ because I multiply both sides by -1} \\ \text{ so } -1<x<0 \implies 0<-x<1 \\ \text{ now you can actually square everything } \\ \implies 0<x^2<1 \\ \text{ so } -1<x<0 \implies 0<x^2<1 \\ \text{ next part is easy } x=0 \implies x^2=0 \\ \\ \text{ now last part } 0<x<5 \implies 0<x<25 \\ \text{ put all of this together } \\ 0<x^2<1 \cup x^2=0 \cup 0<x^2<25 \\ \implies 0 \le x^2<25\]