Limits and Continuity

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]

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 $x=0$.

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 $6$.

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]

There exists a number $c$ in $(a,b)$ such that $h'(c) = 0$.

h is strictly increasing or decreasing over $[a,b]$.

h has at least one real number $c$ in $(a,b)$ where $h(c) = 0$.

Function $h$ must have exactly one root between $a$ and $b$.

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 $g(c)=m$.

There exists exactly one number $c$ in $(-10, 10)$ such that $j(c) = 0$.

Every number in the interval $(-10, 10)$ is a solution to the equation $j(x) = 0$.

There exists at least one number $c$ in $(-10, 10)$ such that $j(c) = 0$.

There is no value that satisfies $j(x) = 0$ within the interval $(-10, 10)$.

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 $[a,b]$.