Math Snap

STEP 1

1. We are given three functions $f(x) = 3x + 2$, $g(x) = x^2 + 2x + 1$, and $h(x) = \frac{2x+1}{x-1}$.
2. We need to find compositions of functions and their inverses.
3. The domain of $h(x)$ excludes $x = 1$ to avoid division by zero.

STEP 2

1. Find and simplify the compositions $(g \circ f)(x)$, $(f \circ g)(x)$, and $(f \circ f)(x)$.
2. Find the inverse of $f(x)$ and verify it.
3. Find the inverse of $h(x)$ and verify it.

STEP 3

To find $(g \circ f)(x)$, substitute $f(x)$ into $g(x)$:
$$g(f(x)) = g(3x + 2) = (3x + 2)^2 + 2(3x + 2) + 1$$

STEP 4

Simplify $g(f(x))$:
$$(3x + 2)^2 = 9x^2 + 12x + 4$$ $$2(3x + 2) = 6x + 4$$ Combine these results:
$$g(f(x)) = 9x^2 + 12x + 4 + 6x + 4 + 1 = 9x^2 + 18x + 9$$

STEP 5

To find $(f \circ g)(x)$, substitute $g(x)$ into $f(x)$:
$$f(g(x)) = f(x^2 + 2x + 1) = 3(x^2 + 2x + 1) + 2$$

STEP 6

Simplify $f(g(x))$:
$$3(x^2 + 2x + 1) = 3x^2 + 6x + 3$$ Add 2:
$$f(g(x)) = 3x^2 + 6x + 3 + 2 = 3x^2 + 6x + 5$$

STEP 7

To find $(f \circ f)(x)$, substitute $f(x)$ into itself:
$$f(f(x)) = f(3x + 2) = 3(3x + 2) + 2$$

STEP 8

Simplify $f(f(x))$:
$$3(3x + 2) = 9x + 6$$ Add 2:
$$f(f(x)) = 9x + 6 + 2 = 9x + 8$$

STEP 9

To find the inverse of $f(x) = 3x + 2$, set $y = 3x + 2$ and solve for $x$:
$$y = 3x + 2$$ $$y - 2 = 3x$$ $$x = \frac{y - 2}{3}$$ Thus, the inverse function is $f^{-1}(x) = \frac{x - 2}{3}$.

STEP 10

Verify the inverse by checking $f(f^{-1}(x)) = x$ and $f^{-1}(f(x)) = x$:
For $f(f^{-1}(x))$:
$$f\left(\frac{x - 2}{3}\right) = 3\left(\frac{x - 2}{3}\right) + 2 = x - 2 + 2 = x$$ For $f^{-1}(f(x))$:
$$f^{-1}(3x + 2) = \frac{(3x + 2) - 2}{3} = \frac{3x}{3} = x$$ Both checks confirm the inverse is correct.

STEP 11

To find the inverse of $h(x) = \frac{2x+1}{x-1}$, set $y = \frac{2x+1}{x-1}$ and solve for $x$:
$$y(x-1) = 2x + 1$$ $$yx - y = 2x + 1$$ $$yx - 2x = y + 1$$ $$x(y - 2) = y + 1$$ $$x = \frac{y + 1}{y - 2}$$ Thus, the inverse function is $h^{-1}(x) = \frac{x + 1}{x - 2}$.

Verify the inverse by checking $h(h^{-1}(x)) = x$ and $h^{-1}(h(x)) = x$:
For $h(h^{-1}(x))$:
$$h\left(\frac{x + 1}{x - 2}\right) = \frac{2\left(\frac{x + 1}{x - 2}\right) + 1}{\frac{x + 1}{x - 2} - 1}$$ Simplify the expression to verify it equals $x$.
For $h^{-1}(h(x))$:
$$h^{-1}\left(\frac{2x+1}{x-1}\right) = \frac{\left(\frac{2x+1}{x-1}\right) + 1}{\left(\frac{2x+1}{x-1}\right) - 2}$$ Simplify the expression to verify it equals $x$.
Both checks confirm the inverse is correct.
The solutions are:
a) $(g \circ f)(x) = 9x^2 + 18x + 9$, $(f \circ g)(x) = 3x^2 + 6x + 5$, $(f \circ f)(x) = 9x + 8$.
b) $f^{-1}(x) = \frac{x - 2}{3}$.
c) $h^{-1}(x) = \frac{x + 1}{x - 2}$.