Find all points of discontinuity of the function $f(t) = \dfrac{1}{t^2 + t - 2}$, where $t = \dfrac{1}{x - 1}$.

Find all points of discontinuity of the function f – Class 12 Mathematics

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f(t) = \dfrac{1}{t^2 + t - 2}

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Appeared in past exams:MHT-CET 2023· Set pcm-2023-05-13-M· 2mexact

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Concept understanding — Continuity of Compositions

Continuity of Composite Functions

Many functions we meet are really one function fed into another: $\sin(x^2)$ is "square, then take sine"; $\sqrt{x^2 + 1}$ is "add one to the square, then take the root." A natural question is: if each piece is continuous, is the combined function continuous? The answer is a reassuring yes, and it saves an enormous amount of work.

The Composition Rule

Important

If $g$ is continuous at $x = a$ , and $f$ is continuous at the point $g(a)$ , then the composite $(f \circ g)(x) = f\big(g(x)\big)$ is continuous at $x = a$ .

Why it works, in plain terms: as $x \to a$ , continuity of $g$ pushes $g(x) \to g(a)$ . Then continuity of $f$ at that landing point pushes $f\big(g(x)\big) \to f\big(g(a)\big)$ . The limit slides cleanly through both functions:

\lim_{x \to a} f\big(g(x)\big) = f\Big(\lim_{x \to a} g(x)\Big) = f\big(g(a)\big).

That last equality is the definition of continuity for $f \circ g$ at $a$ .

Using It in Practice

Most "is this function continuous?" problems become one-liners:

$\sin(x^2)$ — $x^2$ is continuous everywhere and $\sin$ is continuous everywhere, so the composite is continuous for all $x$ .
$\sqrt{x^2 + 1}$ — the inside is continuous and always $\ge 1 > 0$ , and $\sqrt{\ }$ is continuous on $(0, \infty)$ , so the composite is continuous everywhere.
$e^{\cos x}$ — continuous on all of $\mathbb{R}$ , being a composition of two everywhere-continuous functions.

The Trap to Watch

Watch out

The outer function must be continuous at the value $g(a)$ , not merely somewhere. For $f(u) = \sqrt{u}$ with $g(x) = x - 4$ , the composite $\sqrt{x-4}$ needs $g(a) \ge 0$ ; at $a = 2$ , $g(2) = -2$ lands outside the domain of $\sqrt{\ }$ , so continuity there simply doesn't apply. …

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The composite fails where the inner map $t=\frac{1}{x-1}$ is undefined ( $x=1$ ) and where the outer denominator $(t+2)(t-1)$ is zero, i.e. $t=-2\Rightarrow x=\frac12$ and $t=1\Rightarrow x=2$ . Discontinuities: $x=\frac12,\ 1,\ 2$ .

The idea

We have a function of a function: $f(t)=\dfrac{1}{t^2+t-2}$ with $t=\dfrac{1}{x-1}$ . A composite can break for two reasons — the inner function may not be defined, or the outer function may not be defined at the value the inner one produces. We check both.

Step-by-step

1. Where the inner function fails. $t=\dfrac{1}{x-1}$ needs $x-1\ne 0$ , so it is undefined at $x=1$ . The composite is therefore discontinuous at $x=1$ .

2. Where the outer function fails. Factor the denominator:

t^2+t-2=(t+2)(t-1),

which is zero at $t=-2$ and $t=1$ . At these $t$ -values $f$ has vertical asymptotes. …

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Q.Find all points of discontinuity of the function $f(t) = \dfrac{1}{t^2 + t - 2}$ , where $t = \dfrac{1}{x - 1}$ .

Concept understanding — Continuity of Compositions

Continuity of Composite Functions

The Composition Rule

Using It in Practice

The Trap to Watch

The idea

Step-by-step

Method: Discontinuities of a Composite Function Defined via Substitution

Steps

Common Mistakes

Q.Find all points of discontinuity of the function f(t)=1t2+t−2f(t) = \dfrac{1}{t^2 + t - 2}f(t)=t2+t−21​, where t=1x−1t = \dfrac{1}{x - 1}t=x−11​.

The idea

Step-by-step

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Q.Find all points of discontinuity of the function $f(t) = \dfrac{1}{t^2 + t - 2}$ , where $t = \dfrac{1}{x - 1}$ .

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