Hard Quant problem

GRE Quant Problem Solving
1

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Hi Gregmat,

For this question, in the solution theres a part in the algebra section where it says “Finally, use the second equation to bring the n+1 inside the function” and it shows the result of it but im not too sure how it got to the result.

Could someone help explaining to me how to put the equation f(x^n+1)/n+1 = x^n into the second equation f(nx) = nf(x) ?

Thanks in advance :slight_smile:

2

Notice that

n f(x) = f(nx)

Replace n with \frac{1}{n + 1}.

3

If i replace n with 1/ n + 1 into that second equation, isnt it just (1/n+1)(f(x) = f((1/n+1*x)) ?

4

No?

You’re saying that

\frac{1}{n + 1} f(x) = f\left(\frac{1}{x} + 1 \times x\right)

Or am I reading it wrong?

5

You read the left part correctly but the right side is f(1/(n+1) * x) if you plug in 1/(n+1) for n like you told me to do. Is this the right equation to start with?

6

Indeed. And notice that you already (in a prior step) got

f\left(\frac{x^{n + 1}}{n + 1}\right)

This is equal to

f\left(\frac{1}{n + 1} \times x^{n + 1} \right)

Can you use that rule now?

7

After replacing 1/(n+1) into the second equation, on the left side of the equation it gives f(1/(n+1) * x).

It is just x ,not x to the power of anything so how can that turn into x to the power of n + 1?

8

Replace x with x^{n + 1} and see what you get. Remember that if

f(nx) = n f(x)

it’s also the case that

f(nx^n) = nf(x^n)
9

Okay I get that. But even after that what happens to the left side of the equation which was 1/(n+1) * f(x)?

Sorry I think I am confused because there are so many variables.

10

So what do you have?

f\left(\frac{x^{n + 1}}{n + 1}\right) = \frac{1}{n + 1} f(x^{n + 1})

Use the first equation in the question now.

11

The right side of the equation will turn into 1/(n+1) * n * x^n

12

Not quite - take another look.

13

Doesn’t the first equation turn into f(x^n+1) = n * x ^n ?

14

The first equation says that

f(x^n) = n x^{n - 1}

What do you get when you replace n with n + 1?

15

That’s what I did above. Don’t you get f(x^n+1) = nx^n?

16

No. It would be f(x^{n + 1}) = (n + 1) x^n.