I've been asked to learn function compositions. I am lost how to find (f ∘ g)(x)

453 Views Asked by At

I am trying to combine functions. My dad explained that he cannot answer and maybe I should track math stack overflow. I am not sure exactly how to type the questions because the keyboard doesn't have all the math symbols. My online class doesn't allow copy and paste either. I will try.

I am trying to find $(f \circ g)(x)$ where

$$f(x)=\frac{1}{x^2+3},$$

$$g(x)=\sqrt{x-2}.$$

My lesson said that (f o g)(x) (the o looks like a degree sign, but placed down lower. My keyboard will not allow me to type the actual symbol) f(g(x)), but I don't understand how to use this formula to solve the problem, and my lesson isn't explaining very well either.

I'm hoping somebody can help me out with this and I can actually interact with them to figure out how to do this.

(I am in tenth grade, precal algebra. Attached is a screenshot of the problem that I am trying to solve.)

$$ \begin{aligned} \text{1. }&\text {Find }(f \circ g)(x) \text { where } f(x)=\frac{1}{x^{2}+3} \text { and } g(x)=\sqrt{x-2} \text {. }\\ &\text{a. }(f \circ g)(x)=\frac{1}{x-2}\\ &\text{b. }(f \circ g)(x)=\frac{1}{\sqrt{x-2}+3}\\ &\text{c. }(f \circ g)(x)=\frac{1}{x+1}\\ &\text{d. }(f \circ g)(x)=\sqrt{\frac{-2 x^{2}-5}{x^{2}+3}} \end{aligned} $$

2

There are 2 best solutions below

3
On

The notation $(f \circ g)(x)$ means the same thing as $f(g(x))$. We are told that $f(x)=\frac{1}{x^2+3}$ for all values of $x$. We can rewrite this by saying that $f(y)=\frac{1}{y^2+3}$ for all values of $y$. In particular, if $y=g(x)$, then $$ f(y)=f(g(x))=\frac{1}{g(x)^2+3} \, . $$ Since $g(x)=\sqrt{x-2}$, we find that $$ f(g(x))=\frac{1}{\left(\sqrt{x-2}\right)^2+3}=\frac{1}{(x-2)+3}=\frac{1}{x+1} \, . $$ Hence, option (c) is correct. Here is another way to understand this result. The notation $f(x)=\frac{1}{x^2+3}$ just means $$ f(\text{something})=\frac{1}{\text{something}^2+3} \, . $$ In particular, if that "something" equals $g(x)$, then we get that $$ f(g(x))=\frac{1}{g(x)^2+3} \, , $$ and we arrive at the same result as before. Let me know if you have any questions.

0
On

$$f(x)=\frac{1}{x^2+3},$$

$$g(x)=\sqrt{x-2}.$$

$(f\circ g) (x)\ $ means the same as $\ f(g(x))\ = f(\sqrt{x-2}) = \ldots$

Can you continue this working?