Which of the following is true about the function below?
It sounds like a quiz question you’d see on a math test, a coding interview, or even a casual forum thread. ” Is it increasing? Plus, is it even? The moment you spot a mysterious formula, the brain goes into “detective mode.Does it have a fixed point? The short answer: you’ll have to break it down piece by piece.
Most guides skip this. Don't.
Below I’ll walk you through the whole process—what the function actually does, why anyone would care, the step‑by‑step mechanics, the traps most people fall into, and the handful of tricks that actually work. By the end you’ll be able to look at any cryptic expression and quickly pick the statements that are true, false, or somewhere in‑between Turns out it matters..
What Is the Function Really?
First things first: the function we’re talking about isn’t some abstract concept that lives only in a textbook. It’s a concrete rule that takes an input x and spits out a number. In most of the examples that pop up online the expression looks something like this:
The official docs gloss over this. That's a mistake That alone is useful..
[ f(x)=\frac{x^2-4}{x-2} ]
At a glance it feels like a rational function—numerator over denominator, a classic “hole” waiting to be discovered. But you can also run into variations:
- (f(x)=\frac{x^3-8}{x-2})
- (f(x)=\frac{\sin x}{x})
- (f(x)=\frac{e^x-1}{x})
All of them share a common pattern: a polynomial (or something nicely behaved) divided by a linear term that could become zero. Practically speaking, that’s where the “which of the following is true? ” question usually hides its trick Less friction, more output..
The domain and the hidden cancellation
If you plug x = 2 straight into the first version, you get 0/0—an indeterminate form. In practice that means the function isn’t defined at x = 2 unless we simplify first. Factoring the numerator gives:
[ x^2-4 = (x-2)(x+2) ]
Cancel the (x‑2) and you end up with (f(x)=x+2) for every x except 2. So the function is essentially a straight line with a tiny hole at 2. That hole is the key to most “true/false” statements you’ll see Practical, not theoretical..
Why It Matters
You might wonder why anyone would waste time dissecting a simple rational expression. The answer is twofold.
Real‑world relevance
In engineering, physics, and computer graphics you often run into formulas that look innocent but misbehave at certain points. Think of a control system that divides by a sensor reading; a zero reading could crash the whole thing. Knowing where the hidden discontinuities are can save you from a nasty bug.
The official docs gloss over this. That's a mistake.
Test‑taking strategy
On standardized tests, the “which of the following is true?” format is deliberately designed to trip you up. In practice, the statements usually play on the difference between the simplified function and the original definition. If you understand the underlying mechanics, you’ll spot the correct answer in seconds instead of second‑guessing every choice That's the part that actually makes a difference..
How It Works (Step‑by‑Step)
Below is a repeatable workflow you can apply to any function that looks like a fraction with a potential zero denominator.
1. Identify the problematic points
Set the denominator equal to zero and solve for x Turns out it matters..
For our example:
(x-2 = 0 \Rightarrow x = 2)
That’s the only candidate where the function could misbehave Simple, but easy to overlook. Which is the point..
2. Factor the numerator
Look for common factors with the denominator. Practically speaking, if the numerator is a polynomial, use factoring tricks (difference of squares, sum/difference of cubes, etc. ) And it works..
Example:
(x^2-4 = (x-2)(x+2))
3. Cancel common factors—but only for the simplified expression
After canceling, write down the simplified version and explicitly note the restriction But it adds up..
[ f(x)=\begin{cases} x+2, & x\neq 2\[4pt] \text{undefined}, & x = 2 \end{cases} ]
4. Test the statements one by one
Now compare each claim against the simplified rule and the original domain restriction.
| Statement | How to evaluate |
|---|---|
| f is continuous everywhere | False—hole at 2. |
| f(3) = 5 | True—plug 3 into x+2. |
| limₓ→2 f(x) = 4 | True—limit of x+2 as x→2 is 4. |
| f is differentiable at 2 | False—no value defined, so no derivative. |
5. Check for hidden tricks
Sometimes the question includes a statement like “f has a removable discontinuity at x = 2.” That’s a fancy way of saying the hole can be “filled” by redefining f(2)=4. If the statement says “removable discontinuity” it’s true for our example.
Applying the workflow to other common forms
a) Cubic over linear: (f(x)=\frac{x^3-8}{x-2})
- Denominator zero at 2.
- Numerator factors as ((x-2)(x^2+2x+4)).
- Cancel → (f(x)=x^2+2x+4) for (x\neq2).
- Limit as x→2 is (4+4+4=12).
- Statements about continuity at 2 are false; limit‑based statements are true.
b) Trig over linear: (f(x)=\frac{\sin x}{x})
- Denominator zero at 0.
- No algebraic factor, but we know the limit (\lim_{x\to0}\frac{\sin x}{x}=1).
- Function is undefined at 0, but limit exists.
- “Removable discontinuity at 0” is true; “continuous at 0” is false.
c) Exponential over linear: (f(x)=\frac{e^x-1}{x})
Same pattern as the trig case. The limit as x→0 is 1 (by series expansion). Again, removable discontinuity.
Common Mistakes / What Most People Get Wrong
Mistake #1: Assuming cancellation fixes the hole automatically
People often write “after canceling, the function is x+2 everywhere,” forgetting to carry the domain restriction. The hole doesn’t magically disappear; you have to explicitly redefine the function at that point if you want continuity Which is the point..
Mistake #2: Mixing up limit and value
A classic trap: “limₓ→2 f(x)=4, therefore f(2)=4.” Not true unless the function is redefined at 2. Limits tell you what the function approaches, not what it is The details matter here..
Mistake #3: Ignoring the sign of the denominator
When the denominator can be zero at more than one point (e.And g. , (x^2-4) in the denominator), you need to check each root separately. Overlooking one root can make a “true” statement turn out false.
Mistake #4: Forgetting about vertical asymptotes
If the numerator and denominator share no common factor, the point where the denominator is zero becomes a vertical asymptote, not a removable hole. Statements about “removable discontinuity” would be wrong in that case Practical, not theoretical..
Mistake #5: Over‑relying on calculators
Plugging x=2 into a calculator will usually give an error or “undefined.Practically speaking, ” Some graphing tools will automatically fill the hole, leading you to think the function is continuous. Always double‑check analytically.
Practical Tips / What Actually Works
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Write the domain first. Before you do any algebra, note every x that makes the denominator zero. That list is your “danger zone.”
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Factor aggressively. Even if the numerator looks messy, try synthetic division or the Rational Root Theorem. A hidden factor is often the key to the puzzle.
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Use limits as a safety net. If you’re unsure whether a hole is removable, compute (\lim_{x\to a} f(x)). If the limit exists and is finite, you have a removable discontinuity.
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Create a piecewise definition. Explicitly write the simplified rule and the “undefined at a” clause. That visual cue stops you from accidentally claiming continuity The details matter here..
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Test edge cases. Plug numbers just left and right of the problematic point. Graphs help, but a quick mental check (“what’s f(1.9) vs f(2.1)?”) can confirm your algebra.
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Keep a cheat sheet of common factorizations. Difference of squares, sum/difference of cubes, and the quadratic formula pop up far more often than you think.
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When in doubt, differentiate the simplified form. If a statement asks about the derivative at a point where the original function is undefined, you can still talk about the limit of the derivative—provided the hole is removable No workaround needed..
FAQ
Q: Can a function have more than one removable discontinuity?
A: Absolutely. Any factor that cancels with the denominator creates a hole. As an example, (\frac{(x-1)(x-3)}{(x-1)(x-2)}) has removable holes at 1 and a vertical asymptote at 2.
Q: Is a hole the same as a vertical asymptote?
A: No. A hole (removable discontinuity) can be “filled” to make the function continuous. A vertical asymptote blows up to ±∞ and can’t be patched by redefining a single point.
Q: How do I know if a limit exists when the algebraic cancellation isn’t obvious?
A: Use L’Hôpital’s Rule (differentiate numerator and denominator) or series expansions. For (\frac{\sin x}{x}) the series (\sin x = x - x^3/6 + …) makes the limit clear.
Q: What if the denominator is a higher‑degree polynomial?
A: Find all real roots of the denominator. Each root is a candidate for a discontinuity. Then factor the numerator and see which roots cancel That's the part that actually makes a difference..
Q: Does the “removable discontinuity” label change if I redefine the function?
A: Once you assign a value at the hole that matches the limit, the function becomes continuous there. Technically the original function had a removable discontinuity; the new function is continuous.
That’s it. The next time you see a cryptic fraction with a “which of the following is true?” prompt, you’ll know exactly where to look: factor, cancel, note the hole, and then test each claim against the piecewise definition. The rest is just a matter of careful reading—and maybe a quick sketch of the graph for confidence.
Happy problem‑solving!
The analysis requires meticulous attention to detail. Such precision defines mathematical rigor.
Thus, clarity resolves ambiguities Small thing, real impact..
