# Core 3 help

1. Sep 16, 2007

### CathyLou

Hi.

Could someone please help me with the following question? I would really appreciate it as I am really stuck.

Express (x + 4)/(2x^2 + 3x + 1) - 2/(2x + 1) as a single fraction in its simplest form.

I got that [(x + 4)(2x + 1) - 2(2x^2 + 3x + 1)]/(2x^2 + 3x + 1)(2x + 1)

= (-2x^2 + 3x + 2)/(x^2 + 3x + 1)(2x + 1) but I am not sure where to go from here.

Thank you.

Cathy

2. Sep 16, 2007

### HallsofIvy

Staff Emeritus
Wouldn't it make more sense to use the LEAST common denominator? Since x2+ 3x+ 1= (2x+1)(x+1) that IS a common denominator. You don't need to use the "2x+1" terrm twice. You have
$$\frac{x+4}{x^2+ 3x+ 1}- \frac{2(x+1)}{x^2+ 3x+ 1}$$
That should be much simpler.

(In your last formula, you have dropped the "2" in "2x2+ 3x+ 1".)

3. Sep 16, 2007

### CathyLou

Thanks so much for your help.

I understand now.

Cathy

4. Sep 16, 2007

### CathyLou

Hi.

Could someone please help me with the following question?

f(x) = 3 - x^2, x>= 0.

(b) Sketch the graphs of y = f(x) and y = f^-1(x) on the same diagram.

I got that the inverse function of f is (3 - x)^(1/2) but I do cannot figure out how to draw it in comparison to f(x). Could someone please show me.

g(x) = 8/(3 - x), x cannot = 3.

(e) Solve the equation f^-1(x) = g(x).

I have no idea how to even start this last part.

Any help would be really appreciated.

Thank you.

Cathy

5. Sep 16, 2007

### arildno

(e).
So, you have:

$$\sqrt{3-x}=\frac{8}{(3-x)}\to\sqrt{3-x}=\frac{8}{(\sqrt{3-x})^{2}}$$

Does that help?

6. Sep 16, 2007

### CathyLou

Thanks for your help.

I have the answer now.

Cathy

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