Composition of Functions Questions in English

(A) Given $f(x) = x^{2} + 2x - 3$ and $g(x) = x + 1$.
We need to find $x$ such that $f(g(x)) = g(f(x))$.
First,calculate $f(g(x))$:
$f(g(x)) = f(x + 1) = (x + 1)^{2} + 2(x + 1) - 3 = (x^{2} + 2x + 1) + 2x + 2 - 3 = x^{2} + 4x$.
Next,calculate $g(f(x))$:
$g(f(x)) = g(x^{2} + 2x - 3) = (x^{2} + 2x - 3) + 1 = x^{2} + 2x - 2$.
Equating both expressions:
$x^{2} + 4x = x^{2} + 2x - 2$.
Subtracting $x^{2}$ from both sides:
$4x = 2x - 2$.
$4x - 2x = -2$.
$2x = -2$.
$x = -1$.

(A) Given functions are $f(x) = 5 - x^2$ and $g(x) = 3x - 4$.
To find $(f \circ g)(-1)$,we use the definition of composition of functions: $(f \circ g)(x) = f(g(x))$.
First,calculate $g(-1)$:
$g(-1) = 3(-1) - 4 = -3 - 4 = -7$.
Now,substitute this value into $f(x)$:
$(f \circ g)(-1) = f(g(-1)) = f(-7)$.
Using the definition of $f(x)$:
$f(-7) = 5 - (-7)^2 = 5 - 49 = -44$.
Thus,the value of $(f \circ g)(-1)$ is $-44$.

(C) Let $x_1, x_2 \in S$ such that $f(x_1) = f(x_2)$.
Applying $g$ to both sides,we get $g(f(x_1)) = g(f(x_2))$.
This is equivalent to $(g \circ f)(x_1) = (g \circ f)(x_2)$.
Since $g \circ f$ is given as an injective mapping,$(g \circ f)(x_1) = (g \circ f)(x_2)$ implies $x_1 = x_2$.
Since $f(x_1) = f(x_2)$ leads to $x_1 = x_2$,the function $f$ must be injective.
Therefore,$f$ is necessarily injective.

(B) We are given that $g \circ f: S \rightarrow U$ is a surjective (onto) function.
By definition of a surjective function,for every element $z \in U$,there exists at least one element $x \in S$ such that $(g \circ f)(x) = z$.
This can be written as $g(f(x)) = z$.
Let $y = f(x)$. Since $x \in S$ and $f: S \rightarrow T$,it follows that $y \in T$.
Substituting this into the equation,we get $g(y) = z$.
Since for every $z \in U$,we have found an element $y \in T$ such that $g(y) = z$,it follows that $g: T \rightarrow U$ is a surjective function.
However,$f$ does not necessarily have to be surjective because the elements in $T$ that are not in the image of $f$ do not affect the surjectivity of $g \circ f$ as long as the image of $f$ covers enough elements in $T$ to map onto all of $U$ through $g$.

(C) Given $f: R \rightarrow R$ where $f(x) = e^{x}$ and $g: R \rightarrow R$ where $g(x) = x^{2}$.
We calculate the composite function $(g \circ f)(x) = g(f(x)) = g(e^{x}) = (e^{x})^{2} = e^{2x}$.
For $(g \circ f)(x) = e^{2x}$,if $(g \circ f)(x_{1}) = (g \circ f)(x_{2})$,then $e^{2x_{1}} = e^{2x_{2}}$,which implies $2x_{1} = 2x_{2}$,so $x_{1} = x_{2}$. Thus,$g \circ f$ is injective.
However,the range of $g \circ f$ is $(0, \infty)$,which is not equal to the codomain $R$,so $g \circ f$ is not surjective.
For $g(x) = x^{2}$,$g(-1) = 1$ and $g(1) = 1$,so $g$ is not injective. Also,the range of $g$ is $[0, \infty)$,so $g$ is not surjective.
Therefore,$g \circ f$ is injective but $g$ is not bijective.

(A) Given $f(x) = \frac{x}{x+1}$.
Calculating the first few terms:
$f_1(x) = \frac{x}{x+1}$
$f_2(x) = f(f(x)) = \frac{\frac{x}{x+1}}{\frac{x}{x+1} + 1} = \frac{x}{x + x + 1} = \frac{x}{2x+1}$
$f_3(x) = f(f_2(x)) = \frac{\frac{x}{2x+1}}{\frac{x}{2x+1} + 1} = \frac{x}{x + 2x + 1} = \frac{x}{3x+1}$
By induction,$f_n(x) = \frac{x}{nx+1}$.
Evaluating at $x = -2$:
$f_n(-2) = \frac{-2}{n(-2)+1} = \frac{-2}{-2n+1} = \frac{2}{2n-1}$.
The product is $P = f_1(-2) \cdot f_2(-2) \cdot \ldots \cdot f_n(-2) = \prod_{k=1}^{n} \frac{2}{2k-1} = \frac{2^n}{1 \cdot 3 \cdot 5 \cdot \ldots \cdot (2n-1)}$.

(C) Given $g(x) = 3x^{2} + 2x - 3$. First,calculate $g(2)$:
$g(2) = 3(2)^{2} + 2(2) - 3 = 12 + 4 - 3 = 13$.
We need to find $f(g(2)) = f(13)$.
Given $4g(f(x)) = 3x^{2} - 32x + 72$,substitute $g(f(x)) = 3(f(x))^{2} + 2f(x) - 3$:
$4[3(f(x))^{2} + 2f(x) - 3] = 3x^{2} - 32x + 72$
$12(f(x))^{2} + 8f(x) - 12 = 3x^{2} - 32x + 72$
$12(f(x))^{2} + 8f(x) - (3x^{2} - 32x + 84) = 0$.
Using the quadratic formula for $f(x)$:
$f(x) = \frac{-8 \pm \sqrt{64 - 4(12)(-(3x^{2} - 32x + 84))}}{24} = \frac{-8 \pm \sqrt{64 + 48(3x^{2} - 32x + 84)}}{24}$
$f(x) = \frac{-8 \pm \sqrt{144x^{2} - 1536x + 4096}}{24} = \frac{-8 \pm \sqrt{(12x - 64)^{2}}}{24} = \frac{-8 \pm (12x - 64)}{24}$.
Since $f(0) = -3$,we test the signs at $x=0$:
If we take the positive sign: $f(0) = \frac{-8 + (-64)}{24} = -3$ (Correct).
So,$f(x) = \frac{-8 + 12x - 64}{24} = \frac{12x - 72}{24} = \frac{x - 6}{2}$.
Finally,$f(13) = \frac{13 - 6}{2} = \frac{7}{2}$.