Fundamentals of Differentiation

$5$

Undefined

Depends on the value of x

$0$

This condition only ensures continuity of f but not necessarily differentiability.

Existence of this limit implies there are no critical points for f near $x=a$.

This limit guarantees that the second derivative of f exists and is finite.

This limit defines the derivative of $f$ at $x=a$ and confirms its differentiability there.

The graph of the function has an asymptote at $x=c$.

The graph of the function has either a corner, cusp, or vertical tangent line at $x=c$.

The function cannot be integrated in an interval containing $x=c$.

The derivative of the function exists and is zero at $x=c$.

Discontinuous at $x = a$

Not defined at $x = a$

Continuous at $x = a$

Has an asymptote at $x = a$

The derivative $f'(a) = 0$.

The graph of $f(x)$ has a corner or cusp at $x = a$.

The function has a maximum or minimum at $x = a$.

$f(x)$ is continuous at $x = a$.

s(t) has a local minimum at t=2.

There is a gap in the graph of s(t) at t=2.

The graph of s(t) has a vertical tangent at t=2.

s(t) is discontinuous at t=2.

g(x) = \sin(x)

f(x) = x^2

h(x) = |x| - 1

k(x) = e^x

Function has no relative extrema on $(a,b)$

Function has no inflection points on $(a,b)$

Continuous on $(a,b)$

Has sharp turns or corners on $(a,b)$

$k(z)$ is continuous but has a slope of zero at $z=5$.

$k(z)$ has a removable discontinuity at $z=5$.

$k(z)$ is not differentiable at $z=5$.

$k(z)$ has an inflection point at $z=5$.

answer

A function can be continuous without being differentiable, but cannot be differentiable without being continuous.

Continuity guarantees differentiability at that point.

Discontinuities always occur where functions are not differentiable.

A function can be differentiable without being continuous at the same point.