Unit 2 practice test: Exponential and Logarithmic Functions

Inside first: $g(2) = 0$. Then $f(g(2)) = f(0) = 4$.

• Correct: $4$: Right: the inner value is $g(2) = 0$, and $f(0) = 4$.
• $1$: This is $g(f(2)) = g(3) = 1$: the functions were used in the reverse order.
• $0$: This is only the inner value $g(2) = 0$; it still has to go into $f$.
• $3$: This is $f(2) = 3$; the input $2$ must go through $g$ first.

$f(g(3)) = f(1) = 1$ and $g(f(3)) = g(2) = 0$. Changing the order changes the output.

• Correct: $f(g(3)) = 1$ and $g(f(3)) = 0$: Each composite worked from the inside out: $g(3) = 1$ then $f(1) = 1$; $f(3) = 2$ then $g(2) = 0$.
• $f(g(3)) = 0$ and $g(f(3)) = 1$: The two values are traded: $f(g(3))$ starts with $g(3)$, so it is $f(1) = 1$.
• $f(g(3))$ and $g(f(3))$ are both $1$: Composition is not commutative; here $g(f(3)) = g(2) = 0$, not $1$.
• $f(g(3)) = 2$ and $g(f(3)) = 1$: These are $f(3)$ and $g(3)$, single values with no composition.

$g$ gives $0$ only at the input $2$, so $f(x)$ must be $2$. That happens at $x = -1$ and at $x = 3$.

• Correct: $x = -1$ and $x = 3$: $f(-1) = 2$ and $f(3) = 2$, and $g(2) = 0$: two inputs work.
• $x = 2$: $2$ is the value $f(x)$ must take, not $x$ itself: $g(f(2)) = g(3) = 1$.
• $x = -1$: This works, but it misses $x = 3$, where $f(3) = 2$ too.
• $x = 3$: This works, but it misses $x = -1$, where $f(-1) = 2$ too.

Multiplying the input of $\log_4$ by $4$ adds $1$ to the output: the outputs go $-2, -1, 0, 1, 2, 3$.

• Correct: It increases by $1$: Equal factors in the input give equal differences in the output: $+1$ each time.
• It is multiplied by $4$: Multiplying by $4$ is what happens to the inputs, not the outputs.
• It increases by $4$: The outputs go up by $1$, from $0$ to $1$ to $2$, not by $4$.
• It is multiplied by $\frac{1}{4}$: The outputs rise, from $0$ to $1$ to $2$; they are not divided by $4$.

The range of the inverse is the domain of $h$. A base-$4$ logarithm takes only positive inputs, so $k$ has only positive outputs.

• Correct: $(0, \infty)$: Domain of $h$, $x > 0$, becomes the range of $k$.
• $(-\infty, \infty)$: That is the range of $h$, which becomes the domain of $k$.
• $[1, \infty)$: $k(-1) = \frac{1}{4}$, an output less than $1$.
• $(-\infty, 0)$: Powers of $4$ are positive, never negative.

$h(16) = 2$ because $4^2 = 16$, and $h\left(\frac{1}{4}\right) = -1$ because $4^{-1} = \frac{1}{4}$: the outputs are base-$4$ exponents.

• Correct: $h(x) = \log_4 x$: Each output is the exponent of $4$ that gives the input.
• $h(x) = 4^x$: $4^{16}$ is huge; the table pairs $16$ with $2$.
• $h(x) = \log_x 4$: The base is fixed at $4$; the input $x$ is the number we reach.
• $h(x) = \frac{x}{4}$: $\frac{16}{4} = 4$, but the table gives $h(16) = 2$.