# Homework Help: Let f, g and h be functions defined below:

1. Jun 5, 2015

### Jaco Viljoen

1. The problem statement, all variables and given/known data
f(x)=(√x^2-3x+2)/(2x-3),
g(x)=3/(√(x+3)) and
h(x)=(x^2-5x+6)/(x-2)

which of the following are true:
A)Df={x∈ℝ:x≤1 or x≥2}
B)Dg={x∈ℝ:x≥-3}
C)Dh=ℝ

2. Relevant equations

3. The attempt at a solution
I am using substitution here by replacing the x by the parameters specified and slightly further:

Df={x∈ℝ:x≤1 or x≥2}
f(x)=(√(0^2-3(0)+2))/(2(0)-3) (yes this is ok)
f(x)=(√(1^2-3(1)+2))/(2(1)-3) (yes this is ok)
f(x)=(√(-1^2-3(-1)+2))/(2(-1)-3) (yes this is ok)
f(x)=(√(2^2-3(2)+2))/(2(2)-3) (yes this is ok)
f(x)=(√(3^2-3(3)+2))/(232)-3) (yes this is ok)

Dg={x∈ℝ:x≥-3}
g(x)=3/(√(-3+3)) gets 3/0 Undefined?
g(x)=3/(√(-2+3)) (yes this is ok)

Dh=ℝ
h(x)=(x^2-5x+6)/(x-2)
h(x)=(1^2-5(1)+6)/(1-2) (yes this is ok)
h(x)=(2^2-5(2)+6)/(2-2) (yes this is ok)

A and C are true.

2. Jun 5, 2015

### certainly

[EDIT:-however your reasoning for $h(x)$ is wrong. 2^2-2*5+6/(2-2)=0/0, which is indeterminate.
HINT:-factorize the numerator of $h(x)$ to see why your answer is correct nonetheless.]

Last edited: Jun 5, 2015
3. Jun 5, 2015

### Jaco Viljoen

0/0 is just 0 isn't it?

4. Jun 5, 2015

### certainly

They are the same.

5. Jun 5, 2015

### Jaco Viljoen

So why would you say in is undefined?

6. Jun 5, 2015

### certainly

No. 0/0 is indeterminate. The previous post was for a post that I think you deleted?

7. Jun 5, 2015

### Jaco Viljoen

So wouldn't only A be correct/true,

Factorise
h(x)=(2^2-5(2)+6)=(4-10+6)=0
I don't think I understand...

8. Jun 5, 2015

### certainly

C is also correct.
Factorize $x^2-5x+6$, then see if anything changes in $h(x)$ because of that.

9. Jun 5, 2015

### Staff: Mentor

No, the above should be h(1). h(x) = $\frac{x^2 - 5x + 6}{x - 2}$
This (above) should be h(2), which is NOT OK. As certainly said, this results in the indeterminate form 0/0, which should suggest that 2 is not in the domain of h.

10. Jun 5, 2015

### Jaco Viljoen

Mark,
so is it only A?
Certainly agrees with my initial answer, but I am not sure why?
He has asked me to factor as above, I have but still not clear why?

11. Jun 5, 2015

### Staff: Mentor

Write the formula for h(x) in factored form. IOW,
$$h(x) = \frac{(?)(?)}{x - 2}$$

In the factored form it is easier to see that for some value(s) of x, both the numerator and denominator are zero. This means that at any such value, the function h is not defined, so the domain is not all of R (the reals).

12. Jun 5, 2015

### Jaco Viljoen

(x-3)(x-2)
if x is 2 or 3 it will be zero,
but how can 0/0 undefined = ℝ?

13. Jun 5, 2015

### Staff: Mentor

The full answer to my question is: $h(x) = \frac{(x-3)(x-2)}{x - 2}$
h(3) = 0/1 = 0
h(2) is not zero. Since division by zero is not defined, 2 is not in the domain of h.

0/0 is not a number -- it is one of several indeterminate forms. These usually show up as limits, which I'm pretty sure you haven't covered yet.

14. Jun 5, 2015

### certainly

I was a bit confused there about something, Mark is right, the domain is all of $\mathbb{R}$ except 2, so yeah, C is wrong.

15. Jun 5, 2015

### vela

Staff Emeritus
When determining the domain here, you don't need to look at what the numerator is equal to. You simply have to note that the denominator is 0 when x=2, so h isn't defined for x=2. Division by 0 is always undefined.