Function Exercises

Approach:

The domain of a sum is the intersection of the domains of its terms, so restrict the radical term and the logarithmic term separately and then intersect the two conditions.

Steps:

1. For $\frac{1}{\sqrt{4 - x^2}}$, the denominator must be non-zero, so the radicand must be strictly positive: $4 - x^2 > 0$.
2. Solving $x^2 < 4$ gives $-2 < x < 2$.
3. For $\ln(x)$, the argument must be strictly positive: $x > 0$.
4. Intersecting the two conditions, $\{-2 < x < 2\} \cap \{x > 0\}$, leaves $0 < x < 2$.

Answer:

The domain of $f(x)$ is $(0, 2)$.

Approach:

Parity has two halves. First check that the domain is symmetric about the origin, because a function that is not defined on a symmetric domain cannot be even or odd. Then compute $f(-x)$ and compare it with $f(x)$.

Steps:

1. The domain requires $\frac{1-x}{1+x} > 0$. Multiplying by $(1+x)^2 > 0$ (for $x \neq -1$) gives $(1-x)(1+x) > 0$, that is $1 - x^2 > 0$, so the domain is $(-1, 1)$ — symmetric about the origin.
2. Compute $f(-x) = \ln\left(\frac{1 + x}{1 - x}\right)$.
3. Since $\frac{1+x}{1-x} = \left(\frac{1-x}{1+x}\right)^{-1}$, we get $f(-x) = \ln\left(\left(\frac{1-x}{1+x}\right)^{-1}\right) = -\ln\left(\frac{1-x}{1+x}\right) = -f(x)$.
4. Because $f(-x) = -f(x)$ on a symmetric domain, the function is odd.

Answer:

$f(x) = \ln\left(\frac{1 - x}{1 + x}\right)$ is an odd function.

Approach:

Solve the equation for $x$ in terms of $y$, then swap the roles of $x$ and $y$. Finally, the domain of the inverse is the range of the original.

Steps:

1. From $y = e^{2x} + 1$ we get $y - 1 = e^{2x}$.
2. Taking logarithms, $\ln(y - 1) = 2x$, so $x = \frac{1}{2}\ln(y - 1)$.
3. Swapping $x$ and $y$ gives the customary form $y = \frac{1}{2}\ln(x - 1)$.
4. The range of the original function: since $e^{2x} > 0$ for all $x$, we have $e^{2x} + 1 > 1$, so the range is $(1, +\infty)$. That is the domain of the inverse.

Answer:

The inverse is $y = \frac{1}{2}\ln(x - 1)$, with domain $(1, +\infty)$.

Approach:

Compute $f(-x)$ and compare it with $f(x)$. For a polynomial this is a two-line calculation.

Steps:

1. Compute $f(-x) = (-x)^3 + (-x) = -x^3 - x = -(x^3 + x) = -f(x)$.
2. Since $f(-x) = -f(x)$ for every $x$, the function is odd. (Its domain, all of $\mathbb{R}$, is symmetric, so the test is legitimate.)

Answer:

$f(x) = x^3 + x$ is an odd function.

Approach:

Find the period of each term separately, then take the smallest common multiple of the two periods.

Steps:

1. $\sin(2x)$ has period $T_1 = \frac{2\pi}{2} = \pi$.
2. $\cos(3x)$ has period $T_2 = \frac{2\pi}{3}$.
3. Express both in units of $\frac{\pi}{3}$: $\pi = \frac{3\pi}{3}$ and $\frac{2\pi}{3} = \frac{2\pi}{3}$. The smallest positive common multiple is $2\pi$.

Answer:

The period of $f(x) = \sin(2x) + \cos(3x)$ is $2\pi$.

Approach:

Differentiate and read the monotonicity off the sign of the derivative.

Steps:

1. Apply the quotient rule: $f'(x) = \frac{2x \cdot x^3 - (x^2 + 1) \cdot 3x^2}{x^6}$.
2. Simplify the numerator: $2x^4 - 3x^4 - 3x^2 = -x^4 - 3x^2 = -x^2(x^2 + 3)$, so $f'(x) = \frac{-(x^2+3)}{x^4}$.
3. On $(0, +\infty)$ we have $x^2 + 3 > 0$ and $x^4 > 0$, so $f'(x) < 0$ everywhere on that interval.
4. A negative derivative on an interval means the function is strictly decreasing there.

Answer:

$f(x) = \frac{x^2 + 1}{x^3}$ is strictly decreasing on $(0, +\infty)$.

Approach:

The outer function $\ln u$ requires $u > 0$ — strictly positive, not merely non-negative. That stricter requirement is what removes the endpoints here.

Steps:

1. The inner function $\sqrt{4 - x^2}$ is defined when $4 - x^2 \geq 0$, i.e. $-2 \leq x \leq 2$.
2. But the outer function $\ln u$ requires $u > 0$, so we need $\sqrt{4 - x^2} > 0$ strictly.
3. Therefore $4 - x^2 > 0$, that is $-2 < x < 2$. The endpoints $x = \pm 2$ are excluded, because there the logarithm’s argument would be $\ln 0$, which is undefined.

Answer:

The domain of $f(x)$ is $(-2, 2)$.

Approach:

Clear the denominator, collect the terms in $x$, then solve. Do not forget to carry the excluded value over to the domain of the inverse.

Steps:

1. From $y = \frac{3x-2}{x+1}$ we get $y(x+1) = 3x - 2$, that is $yx + y = 3x - 2$.
2. Collect the $x$-terms: $x(y - 3) = -y - 2$, so $x = \frac{y + 2}{3 - y}$, valid for $y \neq 3$.
3. Swapping $x$ and $y$ gives $y = \frac{x + 2}{3 - x}$.
4. The value $y = 3$ can never be produced by the original function (setting $\frac{3x-2}{x+1} = 3$ forces $3x - 2 = 3x + 3$, a contradiction), so the range of the original is $\mathbb{R} \setminus \{3\}$, and that is the domain of the inverse.

Answer:

The inverse is $y = \frac{x + 2}{3 - x}$, with domain $\mathbb{R} \setminus \{3\}$.

Approach:

Differentiate both sides with respect to $x$ — treating $y$ as a function of $x$, so that $\frac{dy}{dx}$ appears — then solve for $\frac{dy}{dx}$ and substitute the point.

Steps:

1. Differentiate $x^2 + xy + y^2 = 3$ with respect to $x$: $2x + \left(y + x\frac{dy}{dx}\right) + 2y\frac{dy}{dx} = 0$.
2. Collect the $\frac{dy}{dx}$ terms: $\frac{dy}{dx}(x + 2y) = -(2x + y)$, so $\frac{dy}{dx} = -\frac{2x + y}{x + 2y}$.
3. Substitute the point $(1, 1)$: $\frac{dy}{dx} = -\frac{2 \cdot 1 + 1}{1 + 2 \cdot 1} = -\frac{3}{3} = -1$.

Answer:

At the point $(1, 1)$ the derivative is $-1$.

Approach:

The argument of the logarithm must be strictly positive, which gives a quadratic inequality to solve.

Steps:

1. Require $x^2 - 4 > 0$.
2. This is the same as $x^2 > 4$, i.e. $|x| > 2$.
3. Splitting into two branches gives $x > 2$ or $x < -2$.

Answer:

The domain of $f(x)$ is $(-\infty, -2) \cup (2, +\infty)$.

Approach:

Compute $f(-x)$ and test both identities. Remember that a single coincidental agreement at one point proves nothing — the identity must hold for all $x$.

Steps:

1. Compute $f(-x) = e^{-x}\cos(-x) = e^{-x}\cos x$, using the evenness of cosine.
2. Test the even identity: $f(-x) = f(x)$ would require $e^{-x} = e^x$, which holds only for $x = 0$. So $f$ is not even.
3. Test the odd identity: $f(-x) = -f(x)$ would require $\cos x = 0$ (after dividing by $e^x \cos x$ when it is non-zero), which holds only at isolated points, not identically. So $f$ is not odd.
4. Neither identity holds, so the function is neither even nor odd.

Answer:

$f(x) = e^x \cos x$ is neither an even function nor an odd function.

Approach:

The arcsine is defined on $[-1, 1]$, so its argument must land in that interval. This gives a linear inequality in $x$.

Steps:

1. Require $-1 \leq \frac{x}{2} \leq 1$.
2. Multiplying through by the positive number $2$ preserves the inequalities: $-2 \leq x \leq 2$.
3. The endpoints are included, because $\arcsin(-1) = -\frac{\pi}{2}$ and $\arcsin(1) = \frac{\pi}{2}$ are perfectly well defined.

Answer:

The domain of $f(x)$ is $[-2, 2]$.

Approach:

Again, one condition per piece, then intersect.

Steps:

1. For $\frac{1}{x-2}$, the denominator must not vanish: $x - 2 \neq 0$, i.e. $x \neq 2$.
2. For $\sqrt{x+1}$, the radicand must be non-negative: $x + 1 \geq 0$, i.e. $x \geq -1$.
3. Intersecting: $x \geq -1$ and $x \neq 2$. The point $x = 2$ is removed from the interval, and the endpoint $x = -1$ is kept because the square root of zero is allowed.

Answer:

The domain of $f(x)$ is $[-1, 2) \cup (2, +\infty)$.

Approach:

A relation is a function exactly when each admissible input $x$ has one and only one output $y$. Test the three relations against that single-valuedness requirement.

Steps:

1. $y = \pm\sqrt{x}$ is not a function: for any $x > 0$ the symbol $\pm$ assigns two values of $y$, namely $\sqrt{x}$ and $-\sqrt{x}$, violating single-valuedness.
2. $y = x^2$ is a function: every real $x$ is assigned exactly one value $y = x^2$.
3. $x^2 + y^2 = 1$ is not a function: for every $x$ with $-1 < x < 1$ the equation gives $y = \pm\sqrt{1 - x^2}$, again two values for one input.

Answer:

Only $y = x^2$ defines a function.