Q.Given the function $f(x) = \dfrac{1}{x + 2}$. Find the points of discontinuity of the composite function $y = f(f(x))$.

Concept: Continuity of Compositions — the composite is discontinuous wherever the inner function is undefined, or wherever the outer function is undefined at the inner function's value.

Step 1: $f(x) = \dfrac{1}{x+2}$ is undefined at $x=-2$.

Step 2: $f(f(x)) = \dfrac{x+2}{2x+5}$ (valid for $x \neq -2$), undefined at $x = -\dfrac52$. …

Setting Up

$f(x)=\dfrac{1}{x+2}$. The composite $f(f(x))$ fails to be defined wherever the inner function $f(x)$ is undefined, or wherever the outer $f$ is undefined at the value $f(x)$ produces. Both must be checked.

Step 1 — Where the inner function is undefined

At $x=-2$, $f(x)$ itself doesn't exist, so $f(f(x))$ can't exist there either.

Step 2 — Compute $f(f(x))$ explicitly

Method: Locating Discontinuities of a Composite Function $f(f(x))$

This method applies whenever you need every point where a composite function built from a given $f$ (applied to itself, or to another function) fails to be defined — it requires checking two distinct sources of discontinuity, not just the simplified formula's own denominator.

Steps

Step 1: Find where the inner function itself is undefined

Any point where $f(x)$ has no value is automatically a point where $f(f(x))$ cannot be evaluated either, since there is nothing to feed into the outer $f$. Record these points first, before doing any algebra.

Step 2: Substitute and simplify the composite expression

Write $f(f(x))$ out explicitly by substituting $f(x)$ into $f$, then simplify algebraically into a single rational expression.

Step 3: Find where the SIMPLIFIED expression's own denominator vanishes

$f(f(x)) = \frac{\text{simplified numerator}}{\text{simplified denominator}}$ …

Common Mistakes

Mistake 1: Only checking the simplified expression's denominator, and missing the inner function's own undefined point

Why it's wrong: simplifying $f(f(x))$ algebraically (combining fractions and cancelling a common factor) is only valid where the original expression was defined in the first place — the simplification can make a genuine discontinuity from the inner function disappear from view. Correct approach: always separately record every point where the inner function is undefined, in addition to solving where the simplified denominator equals zero.

Mistake 2: Not cross-checking where the inner function's output lands on the outer function's own undefined value …