Function Notation and Evaluation: Practice

Square the whole input $x + h$.

$f(x + h) = (x + h)^2 = x^2 + 2xh + h^2$

The middle term $2xh$ is the one most often dropped.

Subtract the values, then divide.

$\frac{(3x + 3h + 2) - (3x + 2)}{h} = \frac{3h}{h} = 3$

The difference quotient of a line is its slope.

Compute each side separately.

$f(2 + 3) = 5^2 = 25, \qquad f(2) + f(3) = 4 + 9 = 13$

Squaring is not additive, so the two are not equal.

Test the through-origin rule $f(x) = 3x$.

$f(a + b) = 3(a + b) = 3a + 3b = f(a) + f(b)$

The others fail; for $f(x) = x + 1$, $f(a) + f(b) = a + b + 2$ but $f(a + b) = a + b + 1$.

Compute each expression.

$f(2x) = (2x)^2 = 4x^2, \qquad 2f(x) = 2x^2$

Since $4x^2 \ne 2x^2$ except at $x = 0$, doubling the input is not doubling the output.

Since $2 \ge 2$, use the second rule.

$f(2) = 3(2) - 1 = 5$

The $x^2$ case needs $x < 2$, so it does not apply at the boundary.

Since $-1 < 2$, use the first rule.

$f(-1) = (-1)^2 = 1$

Wrap the input in parentheses so the square is positive.

Take both with codomain the real numbers. For $x \ne 0$, the fraction cancels to $x$, so the rules agree there.

$f(0) = \tfrac{0}{0} \ \text{is undefined}, \qquad g(0) = 0$

Since $0$ is in $g$'s domain but not $f$'s, the domains differ and the functions are different, even though the codomains match.

The whole input $a + 1$ goes in the denominator.

$f(a + 1) = \frac{1}{a + 1}$

Do not split it into $\tfrac{1}{a} + 1$.

It is the slope of the line through $(x, f(x))$ and $(x + h, f(x + h))$.

$\text{slope} = \frac{f(x + h) - f(x)}{h}$

That secant slope is the average rate of change over the interval.

The original $\dfrac{2xh + h^2}{h}$ reads $\tfrac{0}{0}$ at $h = 0$.

$\frac{2xh + h^2}{h} = 2x + h \quad (h \ne 0)$

Cancelling $h$ is a division, so the two forms agree only for $h \ne 0$.

Set the output equal to $9$ and solve for the input.

$2x - 1 = 9 \ \Rightarrow \ 2x = 10 \ \Rightarrow \ x = 5$

Here you solve for the input that produces the output $9$.