Functions - Function notation

Given the function $f(x) = 3x^2 - 5x + 2$, calculate the value of $f(4)$.

$f(4) = 3(4)^2 - 5(4) + 2$ $f(4) = 3(16) - 20 + 2$ $f(4) = 48 - 20 + 2 = 30$

Substitute the value $x = 4$ into every instance of $x$ in the function expression and simplify using the order of operations.

If $g(x) = 2x + 1$ and $h(x) = x^2$, find the expression for $(h \circ g)(x)$.

$(h \circ g)(x) = h(g(x))$ $(h \circ g)(x) = h(2x + 1)$ $(h \circ g)(x) = (2x + 1)^2$ $(h \circ g)(x) = 4x^2 + 4x + 1$

To find the composite function $h(g(x))$, substitute the entire expression of $g(x)$ into the function $h$ in place of $x$.

Determine the domain and range of the function $f(x) = \sqrt{x - 3} + 5$.

Domain: $x - 3 \ge 0 \implies x \ge 3$ Range: Since $\sqrt{x-3} \ge 0$, then $\sqrt{x-3} + 5 \ge 5 \implies f(x) \ge 5$

For the domain, the value inside the square root must be non-negative. For the range, we consider the minimum possible value of the square root term (which is $0$) and add the constant vertical shift.

1. Find the gradient $m$: $m = \frac{8 - 4}{2 - 0} = \frac{4}{2} = 2$
2. Use the $y$-intercept $(0, 4)$, which means $c = 4$. Thus, $f(x) = 2x + 4$.
3. Evaluate $f(-1)$: $f(-1) = 2(-1) + 4 = -2 + 4 = 2$

The function is established by finding the slope and intercept from the given coordinates. Once the general expression $f(x)$ is known, any input can be substituted to find the corresponding output.

1. Set the function expression equal to 5: $x^2 - 4 = 5$
2. Add 4 to both sides: $x^2 = 9$
3. Take the square root of both sides: $x = \pm \sqrt{9}$
4. Result: $x = 3$ or $x = -3$.

This problem asks for the inputs (domain elements) that map to a specific output (range element). Graphically, this is where the horizontal line $y=5$ intersects the curve $y=x^2-4$.