Graphs of Functions: Additional Practice (Set 1)

The factor $(x + 2)^3$ has power $3$, an odd power, so it behaves like a single factor for the sign.

$(x + 2)^3 = (x + 2)\big((x + 2)^2\big), \quad \text{sign of } (x + 2)$

The sign flips across $x = -2$, so the graph crosses the axis there, even though the multiplicity is greater than one.

Subtract $f$ and evaluate the product at $x = 5$.

$g(5) - f(5) = c(4)(3)(2)(1) = 24c$

At $x = 5$ none of the factors vanish, so the gap is $24c$, not $0$.

Test $x^4 + 3$ by substituting $-x$.

$f(-x) = (-x)^4 + 3 = x^4 + 3 = f(x)$

Since $f(-x) = f(x)$, it is even. The other three are odd.

Use the odd condition $f(-x) = -f(x)$ at $x = 4$.

$f(-4) = -f(4) = -(-9) = 9$

Negating the input negates the output.

An odd function requires $f(-a) = -f(a)$, so from $(2, 5)$ on the graph we get $f(-2) = -f(2) = -5$, which forces the point $(-2, -5)$.

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

Now check the other three against this. The pair $(2, 5)$ and $(2, -5)$ is not even a function, since the input $2$ would carry two outputs and fail the vertical line test from the relations lesson. The pair $(2, 5)$ and $(-2, 5)$ needs $f(-2) = 5$, and $(2, 5)$ and $(-2, 0)$ needs $f(-2) = 0$; both contradict the forced value $f(-2) = -5$. So $(2, 5)$ and $(-2, -5)$ is the only consistent pair.

Factor completely.

$x^5 - x = x(x^4 - 1) = x(x - 1)(x + 1)(x^2 + 1) = 0$

The factor $x^2 + 1$ has no real zero, so the real zeros are $x = 0, 1, -1$, which is $3$.

Factor the denominator: $x^2 - 25 = (x - 5)(x + 5)$, so the original function's domain excludes both $x = 5$ and $x = -5$. Cancel the common factor for the other inputs.

$\frac{x - 5}{(x - 5)(x + 5)} = \frac{1}{x + 5}, \qquad x \ne 5, \ x \ne -5$

The cancelled factor $(x - 5)$ produces the hole at $x = 5$, while the surviving factor $(x + 5)$ makes the denominator zero at $x = -5$, a vertical asymptote.

A negative-leading quartic is $-ax^4 + bx^3 + cx^2 + dx + e$ with $a > 0$. Factor out the highest power.

$-ax^4 + bx^3 + cx^2 + dx + e = x^4\left(-a + \frac{b}{x} + \frac{c}{x^2} + \frac{d}{x^3} + \frac{e}{x^4}\right)$

Past a large enough threshold on $\lvert x \rvert$ the correction terms are together smaller than $a$, so the bracket stays negative, and $x^4 \ge 0$ makes the product large and negative. Both ends fall to large negative values.

Read the factor signs on $-2 < x < 1$: $(x + 2) > 0$, $(x - 1) < 0$, $(x - 4) < 0$.

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

The product is positive on that interval.

Test $x^3 - 4x$ by substituting $-x$.

$f(-x) = (-x)^3 - 4(-x) = -x^3 + 4x = -(x^3 - 4x) = -f(x)$

Since $f(-x) = -f(x)$, it is odd. The others are even, neither, and even.

The sign pattern of $f = x(x - 2)(x + 2)$ across its roots is $-, +, -, +$. The positive intervals are $-2 < x < 0$ (bounded) and $x > 2$ (unbounded).

$x > 2:\ (+)(+)(+) = + \text{, extending forever}$

So the unbounded positive interval is $x > 2$.

Test $g$ directly, using that $f$ is even.

$g(-x) = f(-x) - 3 = f(x) - 3 = g(x)$

Since $g(-x) = g(x)$ for every $x$, $g$ is even no matter which even $f$ you start with.