Math Snap

STEP 1

1. The compositions of functions involve substituting one function into another.
2. The domain of a composition is determined by the domain of the inner function and the resulting domain after substitution.
3. Inverses of functions are found by solving the equation $y = f(x)$ for $x$.

STEP 2

1. Find the compositions of functions.
- Evaluate $f \circ g$.
- Evaluate $g \circ f$.
- Evaluate $f \circ h$.
- Evaluate $g \circ h$.
- Evaluate $h \circ f$.
- Evaluate $h \circ g$.
2. Determine the inverses of functions.
- Find the inverse of $f(x) = 4 - 5x$.
- Find the inverse of $h(x) = x^2 - 3x + 2$.
- Find the inverse of $f(x) = \frac{2x+1}{3-x}$ and determine the domain and range.

STEP 3

Evaluate $f \circ g$.
$$f \circ g(x) = f(g(x)) = f(2-x) = 3 - (2-x)^2$$ Simplify the expression:
$$= 3 - (4 - 4x + x^2) = 3 - 4 + 4x - x^2 = -1 + 4x - x^2$$ The domain of $f \circ g$ is all real numbers, $(-\infty, \infty)$, since $g(x)$ is defined for all $x$.

STEP 4

Evaluate $g \circ f$.
$$g \circ f(x) = g(f(x)) = g(3-x^2) = 2 - (3-x^2)$$ Simplify the expression:
$$= 2 - 3 + x^2 = x^2 - 1$$ The domain of $g \circ f$ is all real numbers, $(-\infty, \infty)$, since $f(x)$ is defined for all $x$.

STEP 5

Evaluate $f \circ h$.
$$f \circ h(x) = f(h(x)) = f\left(\frac{1}{x}\right) = 3 - \left(\frac{1}{x}\right)^2$$ The domain of $f \circ h$ is $(0, \infty)$, since $h(x)$ is defined for $x > 0$.

STEP 6

Evaluate $g \circ h$.
$$g \circ h(x) = g(h(x)) = g\left(\frac{1}{x}\right) = 2 - \frac{1}{x}$$ The domain of $g \circ h$ is $(0, \infty)$, since $h(x)$ is defined for $x > 0$.

STEP 7

Evaluate $h \circ f$.
$$h \circ f(x) = h(f(x)) = h(3-x^2) = \frac{1}{3-x^2}$$ The domain of $h \circ f$ is determined by $3-x^2 > 0$, which simplifies to $-\sqrt{3} < x < \sqrt{3}$.

STEP 8

Evaluate $h \circ g$.
$$h \circ g(x) = h(g(x)) = h(2-x) = \frac{1}{2-x}$$ The domain of $h \circ g$ is $x \neq 2$, or $(-\infty, 2) \cup (2, \infty)$.

STEP 9

Find the inverse of $f(x) = 4 - 5x$.
Set $y = 4 - 5x$ and solve for $x$:
$$y = 4 - 5x$$ $$5x = 4 - y$$ $$x = \frac{4 - y}{5}$$ Thus, the inverse is $f^{-1}(x) = \frac{4 - x}{5}$.

STEP 10

Find the inverse of $h(x) = x^2 - 3x + 2$.
Since the domain is $(2, \infty)$, we solve $y = x^2 - 3x + 2$ for $x$ using the quadratic formula:
$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ For $y = x^2 - 3x + 2$, $a = 1$, $b = -3$, $c = 2 - y$.
$$x = \frac{3 \pm \sqrt{(-3)^2 - 4 \cdot 1 \cdot (2-y)}}{2 \cdot 1}$$ $$x = \frac{3 \pm \sqrt{9 - 8 + 4y}}{2}$$ $$x = \frac{3 \pm \sqrt{1 + 4y}}{2}$$ Since the domain is $(2, \infty)$, we take the positive root:
$$x = \frac{3 + \sqrt{1 + 4y}}{2}$$ Thus, the inverse is $h^{-1}(x) = \frac{3 + \sqrt{1 + 4x}}{2}$.

Find the inverse of $f(x) = \frac{2x+1}{3-x}$.
Set $y = \frac{2x+1}{3-x}$ and solve for $x$:
$$y(3-x) = 2x + 1$$ $$3y - yx = 2x + 1$$ $$3y - 1 = 2x + yx$$ $$3y - 1 = x(2 + y)$$ $$x = \frac{3y - 1}{2 + y}$$ Thus, the inverse is $f^{-1}(x) = \frac{3x - 1}{2 + x}$.
The domain of $f$ is $x \neq 3$, or $(-\infty, 3) \cup (3, \infty)$.
The range of $f$ is all real numbers except the value that makes the denominator zero in the inverse, which is $y \neq -2$.
The domain of $f^{-1}$ is $x \neq -2$, or $(-\infty, -2) \cup (-2, \infty)$.