Function Notation and Evaluation: Additional Practice (Set 2)

Three restrictions apply. You divided by $h$ to reach the simplified form, so $h \ne 0$. And $f$ must be defined at both inputs of the difference quotient, at $x$ and at $x + h$, so neither denominator is zero.

$h \ne 0, \quad x \ne 0, \quad x + h \ne 0$

The last condition is the same as $x \ne -h$. The simplified expression by itself is fine at $h = 0$, where it equals $-\tfrac{1}{x^2}$; the exclusion of $h = 0$ comes from the division that produced it.

Combine over $x(x + h)$, then divide by $h$.

$\frac{1}{x + h} - \frac{1}{x} = \frac{-h}{x(x + h)}, \qquad \div h \Rightarrow \frac{-1}{x(x + h)}$

The $h$ cancels for $h \ne 0$.

Put the whole expression $f(x) = x^2 + 1$ into the slot.

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

As notation this is just substitution; the general operation is studied later.

Test the product rule.

$f(ab) = 7ab, \qquad f(a)\,f(b) = 7a \cdot 7b = 49ab$

These differ in general, for example $a = b = 1$ gives $7$ against $49$, so $f(ab) = f(a)\,f(b)$ fails. The additivity and scaling statements do hold for $f(x) = mx$.

Compute each side.

$f(9 + 16) = \sqrt{25} = 5, \qquad f(9) + f(16) = 3 + 4 = 7$

The square root is not additive either, so the two are not equal.

Expand $f(x + h) = (x + h)^2 + b(x + h)$, then subtract $f(x) = x^2 + bx$.

$f(x + h) - f(x) = (x^2 + 2xh + h^2 + bx + bh) - (x^2 + bx) = 2xh + h^2 + bh$

Now divide by $h$.

$\frac{2xh + h^2 + bh}{h} = 2x + h + b$

The linear coefficient $b$ carries straight through.

Since $2$ satisfies $0 \le x \le 2$, use $x^2$.

$f(2) = 2^2 = 4$

The constant $5$ needs $x > 2$, so it does not apply at $2$.

Factor and cancel for $x \ne 2$: $\dfrac{(x - 2)(x + 2)}{x - 2} = x + 2$.

$f(2) = \tfrac{0}{0} \ \text{undefined}, \qquad g(2) = 4$

The domains differ at $x = 2$, so the functions are different.

Since $3 > 2$, use the constant rule.

$f(3) = 5$

The $x^2$ case only covers $0 \le x \le 2$.

Expand $(x + h)^3 = x^3 + 3x^2 h + 3x h^2 + h^3$ and subtract $x^3$.

$\frac{3x^2 h + 3x h^2 + h^3}{h} = 3x^2 + 3xh + h^2$

Every surviving term had a factor of $h$.

Set the output to $0$ and factor.

$x^2 - 3x = 0 \ \Rightarrow \ x(x - 3) = 0 \ \Rightarrow \ x = 0 \ \text{or} \ x = 3$

Here you solve for the inputs that give output $0$.

The constant $+1$ appears in both $f(x + h)$ and $f(x)$, so it cancels.

$\frac{[(x + h)^2 + 1] - [x^2 + 1]}{h} = \frac{2xh + h^2}{h} = 2x + h$

Constants never affect the difference quotient.