Continuity Theorems: 17 MCQs with answers

Extreme Value Theorem

The Extreme Value Theorem states that a continuous function on a closed interval [a, b] attains both a maximum and a minimum value.

Intermediate Value Theorem

The Intermediate Value Theorem (IVT) states that if f is continuous on [a, b] and f(a) and f(b) have opposite signs, then there exists at least one c in (a, b) such that f(c) = 0.

f(x) = |x|

The absolute value function |x| is continuous everywhere. 1/x is discontinuous at x=0, tan(x) is discontinuous at odd multiples of π/2, and floor(x) is discontinuous at integers.

Continuous at x = c

The composition of continuous functions is continuous: if f is continuous at c and g is continuous at f(c), then g ∘ f is continuous at c.

The function is continuous on the closed interval [a, b]

The Extreme Value Theorem requires the function to be continuous on a closed interval [a, b]. Continuity on an open interval is not sufficient.

False

Continuity does not imply differentiability. For example, f(x) = |x| is continuous on [-1,1] but not differentiable at x = 0.

True

If f is continuous on [a, b] and f(a) and f(b) have opposite signs, then by the IVT there exists c in (a, b) such that f(c) = 0.

3

By the Intermediate Value Theorem, f takes on every value between 2 and 5. Since 3 is between 2 and 5, it must be attained.

Find a delta that works for every epsilon

The epsilon-delta definition requires that for every ε > 0, there exists a δ > 0 such that if |x - c| < δ, then |f(x) - f(c)| < ε. So the strategy is to find a δ (often depending on ε) that works for all ε.

The function is differentiable

Continuity on a closed interval implies boundedness, attainment of max/min (Extreme Value Theorem), and intermediate values (IVT), but it does NOT imply differentiability. For example, |x| is continuous but not differentiable at 0.

Assume ε > 0 is given

The epsilon-delta proof starts with an arbitrary ε > 0, then you must find a δ > 0 that satisfies the condition. So the first step is to let ε > 0 be given.

If a function is differentiable at a point, it is continuous there

Differentiability at a point implies continuity at that point. The converse is not true; continuity does not imply differentiability (e.g., |x| at 0).

To show that a certain value is attained by the function

The IVT is used to show that a continuous function takes on a particular value (like zero) on an interval, which helps prove the existence of solutions to equations.

f(x) = 2 for some x in [0, 2]

By the IVT, since 2 is between f(0)=1 and f(2)=3, there exists some c in (0,2) such that f(c)=2. The other options are not guaranteed.

f(x) = x sin(1/x) for x ≠ 0, f(0) = 0

For f(x) = x sin(1/x), as x → 0, x sin(1/x) → 0 because |sin(1/x)| ≤ 1, so the limit equals f(0) = 0. The other functions have limits that do not exist or do not match the value at 0.

Evaluate the function at 0 and 1 and use IVT

Let f(x) = x^3 - 4x + 2. f(0) = 2 and f(1) = -1, so by the IVT, since 0 is between -1 and 2, there exists c in (0,1) with f(c) = 0.

Apply the Intermediate Value Theorem to h(x) = f(x) - x

To prove a fixed point, define h(x) = f(x) - x. If f is continuous on [0,1] and maps [0,1] into [0,1], then h(0) ≥ 0 and h(1) ≤ 0, so by IVT there exists c with h(c) = 0, i.e., f(c) = c.