Limits and Continuity

It has at least one root in $(-4,6)$.

It has no roots outside of $(-4,6)$.

Its maximum value occurs at $x=-4$ or $x=6$.

It only intersects with x-axis once between $-4$ and $6$.

Because $k(a) \cdot k(b) < 0$ implies a change of sign, so by IVT there's some $c$ where $k(c) = 0$ between $a$ and $b$.

Because if $k(a) \cdot k(b) < 0$, it indicates multiple zeros exist within $[a,b]$.

Because it suggests an increasing or decreasing behavior leading to zero somewhere between $a$ and $b$ without applying IVT.

Because it means both $k(a) = k(b) = 0$ which directly provides two zeros within $[a,b]$.

There exists a number $c$ in the interval $(5, \infty)$ such that $f(c) = -1$.

There exists a number $c$ in the interval $(2, 3)$ such that $f(c) = 0$.

There exists a number $c$ in the interval $(2, 3)$ such that $f(c) = -1$.

There exists a number $c$ in the interval $(3, 5)$ such that $f(c) = 0$.

If $f$ is continuous over $[0, 22]$ and $f(0) = 17$ while $f(22) = -17$, then $f(k) = 17$ for some $k$ in $[0, 22]$.

If $f$ is continuous over $[-9, -6]$ and $f(-9) > 8$ while $f(-6) > 8$, then $f(k) = 8$ for some $k$ in $[-9, -6]$.

If $f$ is continuous over $(-4, 5)$ and $f(-4) > 7$ while $f(5) < 7$, then $f(k) = 7$ for some $k$ in $(-4, 5)$.

If $f$ is uniformly continuous over the entire real line, then applying IVT is useless.

h must be defined from [$x_1,x_2$].

$h'$ must exist everywhere between $x_1$ and $x_2$.

n lies between h($x_1$) and h($x_2$).

h must be continuous from [$x_1,x_2$].

$g''(x)$ is positive for all values of $x$ near zero.

$g'(x)$ is less than zero for all values of $x$ near zero.

$g(x)$ has at least one root in the interval containing zero.

The function $g(x)$ reaches its maximum value at $x=0$.

f(x) = \sin(x) on the interval [0, \pi]

f(x) = x^3 - x^2 + x - 1 on the interval [0, 1]

f(x) = x^2 + 1 on the interval [-1, 1]

f(x) = 2x - 1 on the interval [-1, 1]

f(x) = x^3 - x^2 + x - 1 on the interval [-1, 1]

f(x) = sin(x) on the interval [0, π]

f(x) = 2x - 1 on the interval [-1, 1]

f(x) = e^x on the interval [-∞, ∞]

No value c satisfies $g(c)=m$ within (p,q).

There exists exactly one c in (p,q) such that $g(c)=m$.

For all x in (p,q), $g(x)=m$.

There is at least one c in (p,q) such that $g(c)=m$.

f(x) = e^x on the interval [−∞, ∞]

f(x) = x^2 - 3x + 2 on the interval [0, 3]

f(x) = 1/x on the interval [1, 5]

f(x) = \sin(x) on the interval [0, \pi]