Step-by-step process to find $3f(x)+f(g(x))$ and $4f(g(x))+2f(-3)$ when $f(x)=e^{x^2}-1$ and $g(x) = \sqrt{x^2-1}$ I can't understand my teacher explanation about how to solve this kind of operations with functions (I'm sorry for my english)
He gave us these functions
$$f(x)=e^{x^2}-1 \qquad g(x) = \sqrt{x^2-1}$$
And I need to find those values in these other functions
a) $\quad 3f(x)+f(g(x))$
b) $\quad 4f(g(x))+2f(-3)$
I will not lie, I have no idea how to solve this, I know i need to find the first functions inside the others but not idea how or what steps to follow, can you explain it to me please? I'm begging you, thanks.
Answer to A)
$\quad 3f(x)+f(g(x))$
$\quad 3f(e^{x^2}-1)+f(\sqrt{x^2-1} )$
$\quad 3(e^{x^2}-1)+(e^{x^2}-1) $
$\quad 3e^{x^2}-3+e^{x^2}-1$
 A: You do not really "solve" anything here.
If the value of $x$ is given, you can compute the value of $f(x), g(x)$ etc., right?
Once you know the values of $f(x)$ and $g(x)$, you can easily compute the other expressions involving $f(x)$ and $g(x)$
Let $f(x) = e^{x^2} - 1$ and $g(x) = \sqrt{x^2 - 1}$
For example, if you are to calculate $f(1)$, you need to replace $x$ by $1$ in the expression for $f(x)$
$f(1) = e^{1^2} - 1 = e - 1$ and $g(1) = \sqrt{1^2 - 1} = 0$
Similarly, you replace $x$ by $g(x)$ in the expression for $f(x)$ when you need to evaluate $f(g(x))$
$f(g(x)) = e^{{g(x)}^2} - 1 = e^{x^2 - 1} - 1$
$3f(x) + f(g(x)) = 3(e^{x^2} - 1) + e^{x^2 - 1} - 1$
A: Let $f,g:I \to \mathbb{R}$, when $I\subset \mathbb{R}$we define
$(f+g)(x)=f(x)+g(x)$
$(f \cdot g)(x)=f(x)g(x)$
$(f \circ g)(x)=f(g(x))$ it is first apply $g$ and then $f$ it iff $g(x)\in \text{Dom $f$}$
In your case $f(x)=e^{x^2}-1$ and $g(x)=\sqrt{x^2-1}$ I will help you with $b)$
$$4f(g(x))+2f(-3)=4f(\sqrt{x^2-1})+2(e^{9}-1)$$
$$ =4(e^{x^2-1}-1)+2(e^{9}-1)$$
$$ =4e^{x^2-1}-4+2e^{9}-2$$
$$ \fbox {$ 4e^{x^2-1}+2e^{9}-6 $}$$
You should be able to do the $a)$
Hint
Notice that $f(a)$ is the value of put the number $a$ in the formula given by $f$
$$3f(x)+f(g(x))=3(e^{x^2}-1)+e^{x^2-1}-1$$
$$=3e^{x^2}-3+e^{x^2-1}-1$$
$$\fbox { $3e^{x^2}+e^{x^2-1}-4$ }$$
