What Is a Function?: Practice

The only forbidden input is the one that divides by zero.

$\frac{1}{0} \;\text{is undefined}$

So $0$ is excluded, and every other number is allowed.

A real square root needs its inside to be zero or positive, so a negative input has no real output.

$\sqrt{-4} \;\text{is not a real number}$

So $-4$ is excluded.

Set the rule equal to the target output and solve for the input.

$2x + 1 = 9 \;\Longrightarrow\; 2x = 8 \;\Longrightarrow\; x = 4$

Here you are finding the input, not reading it off.

Every output is a square, and a square is never negative.

$x^2 \ge 0$

The outputs fill the nonnegative side, so the range is all numbers greater than or equal to $0$.

A function needs each input $x$ to give exactly one output $y$. The line $y = 2x - 7$ does this, while the others give two $y$ values for some $x$.

$x = 4 \;\Rightarrow\; y = 2 \text{ or } y = -2 \;\text{ for } x = y^2$

So only $y = 2x - 7$ is a function of $x$.

Substitute $\tfrac{1}{2}$ for $x$.

$f\left(\tfrac{1}{2}\right) = 5 - 2 \cdot \tfrac{1}{2} = 5 - 1 = 4$

Two points on one vertical line share the same input but have different outputs.

$\text{one input} \;\longrightarrow\; \text{two outputs}$

One input with two outputs breaks the definition, so the graph is not a function.

Substitute $3$ for $x$ and divide.

$h(3) = \frac{12}{3} = 4$

The inside of a real square root must be zero or positive.

$x \ge 0$

So the domain is all numbers greater than or equal to $0$.

Evaluate each piece, then add.

$f(0) + f(1) = (0^2 + 1) + (1^2 + 1) = 1 + 2 = 3$

Set the rule equal to $13$ and solve for the input.

$4x - 3 = 13 \;\Longrightarrow\; 4x = 16 \;\Longrightarrow\; x = 4$

The rule forbids one input from having two outputs, but it allows different inputs to share an output.

$3 \mapsto 9 \quad \text{and} \quad (-3) \mapsto 9$

So two different inputs giving the same output is the true statement.