MHB Composite functions and substitutions Quick Question

[ardentmed]

Hey guys,

I have a couple more questions about this problem set I've been working on. I'm doubting some of my answers and I'd appreciate some help.

Question:

For the first one, since f o g = √(√x+5)-2), the domain is [-1,infinity)

Likewise, g o f = √(√x-2)+5) and the domain is [27,infinity) since x-2 >= 25.

Also, since f o f = √(√x-2)-2), the domain is [6,infinity) because √x-2) >= 2 which means x >=6.

Furthermore, since g o g = √(√x+5)+5), the domain is [20,infinity) because √x+5) >= -5 which means x >= 25-5
As for the second question, f o g = 3/(3-2x) via simple substitution. And since x cannot = 3/2, the domain is:

(-infinity, 3/2) u (3/2, infinity)

Please tell me if I'm on the right track.

Thanks again guys.

 

[evinda]

Hey guys,

I have a couple more questions about this problem set I've been working on. I'm doubting some of my answers and I'd appreciate some help.

Question:For the first one, since f o g = √(√x+5)-2), the domain is [-1,infinity)

Likewise, g o f = √(√x-2)+5) and the domain is [27,infinity) since x-2 >= 25.

Also, since f o f = √(√x-2)-2), the domain is [6,infinity) because √x-2) >= 2 which means x >=6.

Furthermore, since g o g = √(√x+5)+5), the domain is [20,infinity) because √x+5) >= -5 which means x >= 25-5
As for the second question, f o g = 3/(3-2x) via simple substitution. And since x cannot = 3/2, the domain is:

(-infinity, 3/2) u (3/2, infinity)

Please tell me if I'm on the right track.

Thanks again guys.

Hi! (Wave)

For the first subquestion,it is like that:

$$f(g(x))=\sqrt{\sqrt{x+5}-2}$$

It must be:

$$x+5 \geq 0 \Rightarrow x \geq -5$$

AND

$$\sqrt{x+5}-2 \geq 0 \Rightarrow \sqrt{x+5} \geq 2 \Rightarrow x+5 \geq 4 \Rightarrow x \geq -1 $$

Therefore,$x \geq \max \{ -1,-5\} \Rightarrow x \geq -1$

So,the domain of $f \circ g$ is $[-1,+\infty)$,as you said!

$$g \circ f=g(f(x))=\sqrt{\sqrt{x-2}+5}$$

It must be:

$$x-2 \geq 0 \Rightarrow x \geq 2$$

AND

$$\sqrt{x-2}+5 \geq 0 \Rightarrow \sqrt{x-2} \geq -5 \text{ ,which is true } \forall x$$

Therefore, $x \geq 2$

So,the domain of $g \circ f$ is $[2,+\infty)$.

The domain of $f \circ f =\sqrt{\sqrt{x-2}-2}$ is $[6,+\infty)$ as you correctly mentioned.

$$g \circ g= \sqrt{\sqrt{x+5}+5}$$

It must be:

$$x+5 \geq 0 \Rightarrow x \geq -5$$

AND

$$\sqrt{x+5}+5 \geq 0 \Rightarrow \sqrt{x+5} \geq -5 \text{ which is true } \forall x$$

Therefore,the domain of $g \circ g$ is $[-5,+\infty)$.

 

[evinda]

For the second subquestion,it is like that:

For the second subquestion:

$$f \circ g=\frac{\frac{3}{x}}{\frac{3}{x}-2}=\frac{3}{3-2x}$$
It must be :

$$x \neq 0 \text{ , and also } 3-2x \neq 0 \Rightarrow x \neq \frac{3}{2}$$

Therefore,the domain is:

$$(-\infty,0) \cup (0,\frac{3}{2}) \cup (\frac{3}{2},+\infty)$$

$$g \circ f=\frac{3}{\frac{x}{x-2}}=\frac{3(x-2)}{x}$$

It must be:
$$x \neq 2 \text{ and } x \neq 0$$

Therefore,the domain is:
$$(-\infty,0) \cup (0,2) \cup (2,+\infty)$$