14.
I am given a function 𝑓(π‘₯) and told that 𝑓 βˆ’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 π‘₯! = 0 for some value π‘₯! in the interval βˆ’1 < π‘₯! < +10?
a. Monotonic
b. Continuous
c. Differentiable
d. Quadratic

Continuity guarantees a root exists in (-1, 10) via the Intermediate Value Theorem (IVT).

Problem Analysis

Given f(βˆ’1) > 0 and f(10) < 0, the Intermediate Value Theorem (IVT) states that if
f is continuous on [βˆ’1, 10], then there exists some
xβ‚€ ∈ (βˆ’1, 10) such that f(xβ‚€) = 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) = 1 for x < 0 and
    f(x) = βˆ’1 for x β‰₯ 0 is monotonic decreasing but discontinuous at 0.
    It satisfies f(βˆ’1) = 1 > 0 and f(10) = βˆ’1 < 0, yet no root exists.
  • Continuous: Necessary and sufficient. IVT directly applies: since f is
    continuous on [βˆ’1,10] and 0 lies between f(βˆ’1) and f(10),
    a root must exist.
  • Differentiable: Unnecessary. Continuity alone suffices; differentiability adds smoothness but
    is not required for IVT. For example, the absolute value function f(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.

Β 

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