Explanation

The problem provides the function \(f(x, y) = x^2y\) and also the fact that when a and b are plugged in for x and y, the answer is 6. In other words:

\(f(x, y) = x^2y\)

\(f(a, b) = a^2b = 6\)

The problem asks for the value of f(2a,4b). First, plug 2a in for x and 4b in for y:

\(f(2a, 4b) = (2a)^2(4b)\)

\(f(2a, 4b) = 4a^2(4b)\)

\(f(2a, 4b) = 16a^2b\)

The problem already provides the value for the variables: a2b = 6. Therefore, \(16a^2b = 16(6) = 96\).

Given that \(f(x, y) = x^2y\) and \(f(a, b) = 6\) and we need to find the value of \(f(2a, 4b)\)

To find f(a, b) we need to compare what is inside the bracket in f(a, b) and f(x, y)
=> We need to substitute x with a and y with b in \(f(x, y) = x^2y\) to get the value of f(a, b)

=> \(f(a, b) = a^2b\) = 6 given

Similarly, \(f(2a, 4b)\) = \((2a)^2*(4b)\) = \(4a^2 * 4b\) = 16 * \(a^2b\) = 16*6 = 96

So, Answer will be 96
Hope it helps!

Watch the following video to learn the Basics of Functions and Custom Characters

Hello from the GRE Prep Club BumpBot!

Thanks to another GRE Prep Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.