Domain traps and the graphical check

What you will be able to do

• find the domain of an expression with several inverse functions, including the case of a single point;
• find the range of a sum of inverse functions that rise or fall together;
• avoid the usual traps, and check an answer with a graph or a numerical scan.

The domain: every inverse function brings a condition

• $\sin^{-1}u$ and $\cos^{-1}u$ need $-1 \le u \le 1$.
• $\sec^{-1}u$ and $\operatorname{cosec}^{-1}u$ need $|u| \ge 1$.
• $\tan^{-1}u$ and $\cot^{-1}u$ need nothing.
• A square root needs a non-negative inside, and a denominator must not be $0$.

All of these must hold at the same time, so the domain is where all the intervals overlap.

Example. Find the domain of $f(x) = \sin^{-1}(2x - 1) + \cos^{-1}(3 - 2x)$. The first term needs $-1 \le 2x - 1 \le 1$, which gives $0 \le x \le 1$. The second needs $-1 \le 3 - 2x \le 1$, which gives $1 \le x \le 2$. The two intervals overlap only at $x = 1$. So the domain is the single point $x = 1$, and $f(1) = \sin^{-1}1 + \cos^{-1}1 = \dfrac{\pi}{2}$.

The range

If every part of an expression rises as $x$ rises, the whole expression rises. Then its smallest value is at the left end of the domain, and its largest value is at the right end. The same is true if every part falls.

Example. Let $f(x) = \sin^{-1}x + \tan^{-1}x$ on $[-1, 1]$. Both parts rise. So the least value is $f(-1) = -\dfrac{\pi}{2} - \dfrac{\pi}{4} = -\dfrac{3\pi}{4}$ and the greatest is $f(1) = \dfrac{3\pi}{4}$. The range is $\left[-\dfrac{3\pi}{4}, \dfrac{3\pi}{4}\right]$.

If one part rises and the other falls, look for a way to combine them first. For example, $\sin^{-1}x + \cos^{-1}x$ is the constant $\dfrac{\pi}{2}$.

The usual traps

• Using an identity outside its domain. $\sin^{-1}x + \cos^{-1}x = \dfrac{\pi}{2}$ is true only for $|x| \le 1$.
• Missing a hidden domain. In $\sin^{-1}x + \sin^{-1}(1 - x)$ the second term also limits $x$, to $[0, 2]$.
• False roots from taking a sine or a tangent.
• A range in the wrong order. For a falling function the least value is at the right end.

The graphical check

Draw the left side and the right side of an equation. A solution is a crossing. The graph also shows where the curves exist, so it shows the domain, and it shows when a root of the algebra is not a real crossing. On the computer you can scan the domain with a fine grid and look for the points where the two sides agree.

A worked example

Solve $\sin^{-1}x + \sin^{-1}(1 - x) = \cos^{-1}x$.

1. The domain. We need $-1 \le x \le 1$ and $-1 \le 1 - x \le 1$, so $0 \le x \le 1$.
2. Isolate. $\sin^{-1}(1 - x) = \cos^{-1}x - \sin^{-1}x = \dfrac{\pi}{2} - 2\sin^{-1}x$.
3. Take the sine. $1 - x = \cos\left(2\sin^{-1}x\right) = 1 - 2x^2$. So $2x^2 = x$, and $x = 0$ or $x = \dfrac{1}{2}$.
4. Check. At $x = 0$: $0 + \sin^{-1}1 = \dfrac{\pi}{2} = \cos^{-1}0$. At $x = \dfrac{1}{2}$: $\dfrac{\pi}{6} + \dfrac{\pi}{6} = \dfrac{\pi}{3} = \cos^{-1}\dfrac{1}{2}$. Both are real solutions.

The scan in the program below finds the same two points.

Practise with new numbers

Try it in Python

The program finds the domain of the equation above and then lists the places where the two sides agree.

Common mistakes

• Checking only one inverse function for the domain. All of them must be true together.
• Reading the range of a sum from the ranges of its parts. The sum of two ranges is only a bound, unless the parts move together.
• Not stating the domain of the answer. A solution outside the domain is not a solution.
• Trusting a root without a check. One substitution into the original equation takes a minute.

Remember these

• The domain is the overlap of the conditions of every inverse function. It may be one point.
• If all parts rise, or all fall, the range runs between the values at the two ends of the domain.
• $\sin^{-1}x + \cos^{-1}x$ is constant, equal to $\dfrac{\pi}{2}$.
• A graph or a scan shows the crossings, and it shows whether a root is real.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Find the domain of $\sin^{-1}(2x - 1)$.

2. What is the value of $\sin^{-1}(x - 1) + \cos^{-1}(x - 1)$ for every $x$ in its domain? What is the domain?

3. Find the range of $\sin^{-1}x + \tan^{-1}x$ on $[-1, 1]$.

4. A student says that $x = -1$ solves $\sin^{-1}x + \sin^{-1}(1 - x) = \cos^{-1}x$. Is that right?

5. Use a sketch of $y = \sin^{-1}x$ and $y = x$ to say how many real solutions $\sin^{-1}x = x$ has.

References

• NCERT, Mathematics Textbook for Class XII, Part I, Chapter 2: Inverse Trigonometric Functions.
• OpenStax, Algebra and Trigonometry 2e, Section 8.3: Inverse Trigonometric Functions.
• OpenStax, Calculus Volume 1, Section 1.4: Inverse Functions.