Graphs of Functions: Practice

For an even function, negating the input leaves the output unchanged.

$f(-3) = f(3) = 7 \Rightarrow (-3, 7)$

Even symmetry reflects the point across the y-axis, keeping the height.

Read the factors carefully. For $x > 5$, the factor $(5 - x)$ is negative and $(x + 1)$ is positive.

$(-)(+) = -$

The product is negative, so watch the leading sign of a factor written as $(5 - x)$ rather than $(x - 5)$.

The domain excludes $x = 0$. For every other input, cancel the common factor $x$.

$\frac{x^3 - x}{x} = \frac{x(x^2 - 1)}{x} = x^2 - 1, \qquad x \ne 0$

Evaluate the simplified rule at the excluded input: $0^2 - 1 = -1$, so the hole is at $(0, -1)$.

Read the factor signs on $1 < x < 3$: $(x - 1) > 0$ and $(x - 3) < 0$.

$(+)(-) = -$

The product is negative, so the graph is below the axis between the roots.

Evenness would require $f(-x) = f(x)$ at every input, so a single mismatched pair disproves it.

$f(-1) = (-1)^2 + (-1) = 0, \qquad f(1) = 1^2 + 1 = 2$

Since $0 \ne 2$, the function is not even. The even degree of the leading term is not enough.

Decide touch versus cross by the multiplicity of each factor.

$x^2 \text{ even power} \Rightarrow \text{touch at } 0, \qquad (x - 3) \text{ odd power} \Rightarrow \text{cross at } 3$

The graph touches at $x = 0$ and crosses at $x = 3$.

The sign pattern of $f$ across the roots is $-, +, -, +$. The negative intervals are $x < -2$ and $0 < x < 2$.

$x < -2 \text{ (negative, infinite)}, \qquad 0 < x < 2 \text{ (negative, bounded)}$

Only $0 < x < 2$ is negative and of finite length.

A positive-leading cubic is $ax^3 + bx^2 + cx + d$ with $a > 0$. Factor out the highest power.

$ax^3 + bx^2 + cx + d = x^3\left(a + \frac{b}{x} + \frac{c}{x^2} + \frac{d}{x^3}\right)$

Past a large enough threshold on $\lvert x \rvert$ the correction terms are together smaller than $a$, so the bracket stays positive and the sign of the cubic matches the sign of $x^3$: negative for $x < 0$ and positive for $x > 0$. The graph falls on the left and rises on the right.

Factor top and bottom: $\dfrac{(x - 3)(x + 3)}{x(x - 3)}$. The factor $(x - 3)$ cancels, so $x = 3$ gives a hole, while $x = 0$ gives a vertical asymptote (it does not cancel).

$\frac{x + 3}{x}\Big|_{x = 3} = \frac{6}{3} = 2$

The hole is at $(3, 2)$.

Rotational symmetry about the origin is exactly oddness, so test $f(-x) = -f(x)$.

$f(-x) = (-x)^3 - (-x) = -x^3 + x = -(x^3 - x) = -f(x)$

The other three are even, so only $x^3 - x$ has $180^\circ$ symmetry.

The table suggests $x^3$, but it cannot fix a value off the samples. For any $c$,

$g(x) = x^3 + c(x - 1)(x - 2)(x - 3)$

matches all three rows, and $g(0) = -6c$ can be any number. So $f(0)$ is not determined.

First find the sign of the product, then apply the leading minus. On $1 < x < 2$, $(x - 1) > 0$ and $(x - 2) < 0$.

$-(+)(-) = -(-) = +$

The leading negative flips the sign, so $f$ is positive between the roots.