Function Notation and Evaluation: Practice

Substitute $-2$, keeping it in parentheses so the signs stay correct.

$f(-2) = (-2)^2 - 3(-2) = 4 + 6 = 10$

Both terms turn out positive.

Replace $x$ with $-x$ everywhere.

$f(-x) = 4(-x) - 1 = -4x - 1$

Only the $x$ term changes sign; the constant $-1$ stays.

The whole input $2x$ goes in the slot and gets squared.

$f(2x) = (2x)^2 + 1 = 4x^2 + 1$

The $2$ is squared along with the $x$.

Substitute $t$ for $x$ in the same rule.

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

Renaming the input variable does not change the rule.

Since $5 \ge 0$, use the second rule.

$f(5) = 5 + 3 = 8$

The first rule needs $x < 0$, so it does not apply.

Substitute $0$ for $x$.

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

An input of $0$ does not force the output to be $0$.

Substitute $1$ for $x$.

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

The result is negative.

The whole input $x + 2$ is multiplied by $3$.

$f(x + 2) = 3(x + 2) = 3x + 6$

Distribute the $3$ across both terms.

Replace $x$ with $x + h$.

$f(x + h) = (x + h) - 4 = x + h - 4$

The $h$ is carried along; it does not disappear.

Substitute $3$ for $x$.

$f(3) = 3^2 + 2(3) = 9 + 6 = 15$

Evaluate each term, then add.

Both rules square the input and add $1$, on the same domain.

$f(5) = 26 = g(5)$

The input letter is only a placeholder, so the two are the same function.

Substitute $-1$ for $x$.

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

Multiply first, then subtract.