Functions | Algebra | Quantitative Aptitude

Let’s discuss how to handle functions today. People usually perceive functions as an advanced topic mainly because of the notation. But actually, the function questions are very simplistic and can be solved with a simple process. If we ask you the value of $5x^3$ where $x = 3$ , would you be worried about what to do? We assume you won’t be. Then there should be no problem with “given $f(x)\ =\ 5x^3$ , what is the value of $f(3)$ ?”

Just keep a few simple things in mind:

– $f(x)$ = …. will be followed by an expression in $x$ . This is the core of your calculations. You can turn a blind eye to $f(x)$ – just focus on the expression. For example: $f(x)\ =\frac{(x^2+1)}{5x}$ . Keep your eye on $\frac{(x^2+1)}{5x}$ .

– When faced with “what is $f(a)$ ?” all you have to do is recall the expression given and put $x = a$ in that. It doesn’t matter what $a$ is – wherever you have $x$ , just put ‘ $a$ ’ there. For example: if $f(x)\ =\ \frac{(x^2+1)}{5x}$ , what is $f(5x^3)$ ? Don’t get confused. Here, $a\ =\ 5x^3$ Look for $x$ in the expression and put $5x^3$ in its place.
$f(5x^3)\ =\ \frac{((5x^3)^2+1)}{5(5x^3)}$
If you seem to be getting lost in too many exponents, terms etc, in place of $x$ , put $5x^3$ and put brackets around it as done above. Then simplify by opening the brackets.
$f(5x^3)\ =\ \frac{(25x^6+1)}{25x^3}$

– When we are given that $f(a) = B$ , put $x = a$ in the expression and equate the whole expression to $B$ . For example: $f(x)\ =\ \frac{(x^2+1)}{5x}$ , given that $f(a)\ =\ \frac{2/5}$ , what is the value of $a$ ?
We know how to find $f(a)$ . It is simply $\frac{(a^2+1)}{5a}$ . We are given that this is $\frac{2}{5}$ .
$\frac{(a^2+1)}{5a}\ =\ \frac{2}{5}$
$a^2\ +\ 1\ =\ 2a$
$a^2\ - 2a + 1 = 0$
$(a\ - 1)^2 = 0$
$a = 1$
That is pretty much all you need. Let’s look at a GMAT Prep question on functions.

Question: For which of the following functions ${f}$ is ${f(x) = f(1-x)}$ for all ${x}$ ?
(A) ${f(x) = 1 - x}$
(B) ${f(x)\ =\ 1\ - x^2}$
(C) ${f(x)\ =\ x^2\ - (1 - x)^2}$
(D) ${f(x)\ =\ x^2\times(1\ - x)^2}$
(E) ${f(x)\ =\ \frac{x}{(1\ - x)}}$

Solution: What does this mean: $f(x) = f(1-x)$ ? It means that given a certain expression in $x$ called $f(x)$ , for which function will that be the same as $f(1-x)$ i.e. when you substitute $x$ by $(1-x)$ , which expression will stay the same? Let’s look at each option:

(A) $f(x) = 1 - x$
Substitute $(1 - x)$ in place of $x$ to see what $f(1 - x)$ looks like.
$f(1 - x) = 1 - (1 - x)$
$f(1 - x) = x$
$f(x)$ is not the same as $f(1-x)$ here. Ignore this option.

(B) $f(x)\ =\ 1\ - x^2$
Substitute $(1 - x)$ in place of $x$ to see what $f(1 - x)$ looks like.
$f(1\ - x) = 1 - (1 - x)^2$
$f(1\ - x) = 2x - x^2$
$f(x)$ is not the same as $f(1-x)$ here. Ignore this option.

(C) $f(x)\ =\ x^2\ - (1 - x)^2$
Substitute $(1 - x)$ in place of $x$ to see what $f(1 - x)$ looks like.
$f(1\ - x) = (1 - x)^2 - (1 - (1-x))^2$
$f(1\ - x) = (1 - x)^2 - x^2$
$f(1\ - x) = -x^2 + (1 - x)^2$
$f(x)$ is not the same as $f(1-x)$ here. Ignore this option.

(D) $f(x)\ =\ x^2\times(1\ - x)^2$
Substitute $(1 - x)$ in place of $x$ to see what $f(1 - x)$ looks like.
$f(1\ - x) = (1 - x)^2 \times (1 - (1 - x))^2$
$f(1\ - x) = (1 - x)^2 \times x^2$
$f(1\ - x) = x^2 \times (1 - x)^2$
Note that here, $f(x) = f(1 - x)$ , so this must be our answer. Still, let’s take a look at (E) as well for practice.

(E) $f(x)\ =\ \frac{x}{(1\ - x)}$
Substitute $(1 - x)$ in place of $x$ to see what $f(1 - x)$ looks like.
$f(1\ - x) = \frac{(1 - x)}{(1 - (1-x))}$
$f(1\ - x) = \frac{(1 - x)}{x}$
$f(x)$ is not the same as $f(1-x)$ . Ignore this option.

Answer (D)
A cursory look back at the solution might make you feel that it involves some complicated manipulations but we hope you do see that it is anything but complicated. Now there are some other ways of handling this question too. If you are comfortable with the above, continue with the rest of the post.

Method 2: Number Plugging

We want the expression for which $f(x) = f(1 - x)$ for ALL values of $x$ . So no matter what value we give $x$ , $f(x)$ should be same as $f(1 - x)$ .
Say, if $x = 0$ , for which function is $f(x) = f(1 -x )$ ? i.e. for which function is $f(0) = f(1)$

(A) $f(x) = 1 - x$
$f(0) = 1$ and $f(1) = 0$ . Not equal.

(B) $f(x)\ =\ 1\ - x^2$
$f(0) = 1$ and $f(1) = 0$ . Not equal.

(C) $f(x)\ =\ x^2\ - (1 - x)^2$
$f(0) = -1$ and $f(1) = 1$ . Not equal.

(D) $f(x)\ =\ x^2\times(1\ - x)^2$
$f(0) = 0$ and $f(1) = 0$ . Equal. But when using number plugging, you need to check all options because multiple options could give you equal values. In that case, you would need to try for another value of $x$ .

(E) $f(x)\ =\ \frac{x}{(1\ - x)}$
$f(0) = 0$ and $f(1)$ is not defined. Just to be sure, say $x = -1$ .
$f(-1)\ =\ \frac{-1}{2}$ and $f(2) = -2$ . Not equal.
Answer (D)

Method 3: Intuitive Approach
Try to first focus on the options where terms are added/multiplied rather than subtracted/divided. They are more symmetrical and a substitution may not change the expression. Intuitively, we should check (D) first since it involves multiplication of the terms.

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