14.
I am given a function 𝑓(𝑥) and told that 𝑓 −1 > 0, 𝑓 +10 < 0. Which of the
following properties of the function is necessary and sufficient to conclude that the
function has a root 𝑓 𝑥! = 0 for some value 𝑥! in the interval −1 < 𝑥! < +10?
a. Monotonic
b. Continuous
c. Differentiable
d. Quadratic
(-1, 10) via the Intermediate Value Theorem (IVT). Problem Analysis
Given f(−1) > 0 and f(10) < 0, the Intermediate Value Theorem (IVT) states that if
f is continuous on [−1, 10], then there exists some
x₀ ∈ (−1, 10) such that f(x₀) = 0.
The sign change across zero ensures that the function must cross the x-axis on a connected interval.
Option Breakdown
- Monotonic: Insufficient alone. A step function like
f(x) = 1forx < 0and
f(x) = −1forx ≥ 0is monotonic decreasing but discontinuous at 0.
It satisfiesf(−1) = 1 > 0andf(10) = −1 < 0, yet no root exists. - Continuous: Necessary and sufficient. IVT directly applies: since
fis
continuous on[−1,10]and 0 lies betweenf(−1)andf(10),
a root must exist. - Differentiable: Unnecessary. Continuity alone suffices; differentiability adds smoothness but
is not required for IVT. For example, the absolute value functionf(x) = |x|is continuous but
not differentiable at 0, and IVT still holds. - Quadratic: Irrelevant. Quadratic functions may have roots, but without solving, there is
no general sign-change guarantee. Therefore, this condition is neither necessary nor sufficient.
Correct Answer
Answer: (b) Continuous.
