Algebra - Composite and inverse functions

Given $f(x) = 3x - 2$ and $g(x) = x^2 + 1$, find the expression for $f(g(x))$.

$f(g(x)) = 3(x^2 + 1) - 2$ $f(g(x)) = 3x^2 + 3 - 2$ $f(g(x)) = 3x^2 + 1$

To find the composite function, we substitute the entire expression of $g(x)$ into every instance of $x$ in $f(x)$.

Find the inverse function $f^{-1}(x)$ for $f(x) = \frac{2x + 5}{3}$.

Step 1: Write as $y$: $y = \frac{2x + 5}{3}$ Step 2: Swap $x$ and $y$: $x = \frac{2y + 5}{3}$ Step 3: Solve for $y$: $3x = 2y + 5$ $3x - 5 = 2y$ $y = \frac{3x - 5}{2}$ Therefore, $f^{-1}(x) = \frac{3x - 5}{2}$

To find the inverse, we swap the roles of the input and output variables and rearrange the equation to make the new $y$ the subject.

If $h(x) = 5x - 4$, find the value of $x$ such that $h^{-1}(x) = 2$.

If $h^{-1}(x) = 2$, then by the property of inverses: $x = h(2)$ Substitute $2$ into the original function: $x = 5(2) - 4$ $x = 10 - 4$ $x = 6$

Using the property that if $f^{-1}(a) = b$, then $f(b) = a$ allows us to solve for $x$ without finding the inverse expression first.

1. Find the expression: $(g \circ f)(x) = g(f(x)) = g(x + 3) = (x + 3)^2$.
2. Substitute $x = 2$: $(g \circ f)(2) = (2 + 3)^2 = 5^2 = 25$.

The inner function $f$ adds 3 to the input. The outer function $g$ squares the result. By applying them in sequence, we transform 2 into 5, then 5 into 25.

1. Set $f(x) = f^{-1}(x) \implies \sqrt{x} = x^2$.
2. Square both sides: $x = x^4 \implies x^4 - x = 0$.
3. Factor: $x(x^3 - 1) = 0 \implies x = 0$ or $x = 1$.
4. Since $x \ge 0$, the intersection points are $(0,0)$ and $(1,1)$.

Inverse functions are reflections over $y=x$. Since the square root and squaring functions are inverses (for positive $x$), they intersect at points where $x = y$, specifically at $(0,0)$ and $(1,1)$.