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Re: x(4 – x)
[#permalink]
16 Nov 2021, 10:39
16 Nov 2021, 10:39
1
Scorplong wrote:
Niraj137
your answer only proves that f(x) < 4 for x in (0, 1) U (3, 4), it does not say anything about the value of x in [1, 3]. And unfortunately since the maximum of f(x) occurs at x=2, your answer does not prove a bound on the value of f.
Furthermore, even if the maximum of f occurred in the interval you discuss, you would still need to prove the maximum occurs in that interval. You can't just assume it does, otherwise you may assume incorrectly.
your answer only proves that f(x) < 4 for x in (0, 1) U (3, 4), it does not say anything about the value of x in [1, 3]. And unfortunately since the maximum of f(x) occurs at x=2, your answer does not prove a bound on the value of f.
Furthermore, even if the maximum of f occurred in the interval you discuss, you would still need to prove the maximum occurs in that interval. You can't just assume it does, otherwise you may assume incorrectly.
Well I could take that approach but is derivatives really included in GRE ? I can understand your reasoning completely and I do admit my mistake. But if not for derivatives how is one supposed to know what the maximum value for x(4-x) comes at ? If you know of any such method, I would like to learn it!
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x(4 – x)
[#permalink]
16 Nov 2021, 10:51
16 Nov 2021, 10:51
2
Niraj137 wrote:
Well I could take that approach but is derivatives really included in GRE ? I can understand your reasoning completely and I do admit my mistake. But if not for derivatives how is one supposed to know what the maximum value for x(4-x) comes at ? If you know of any such method, I would like to learn it!
To answer your first question, the GRE syllabus does not require students to know anything about calculus/derivatives.
However, the GRE does expect students to know a few things about parabolas.
For example, the graph of the equation y = x(4-x) is a parabola.
If we expand and rearrange the terms we get the equation: y = -x² + 4x
Since the coefficient of x² is NEGATIVE, we know that the problem opens downward, which means the maximum value of -x² + 4x will occur at the parabola's vertex.
So, once we have the equation y = -x² + 4x, we can use the technique known as "completing the square" to rewrite the equation as y = -1(x - 2)² + 4
Please see my post (https://gre.myprepclub.com/forum/x-4-x-5508.html#p9771) for a step-by-step approach to completing the square.
Re: x(4 – x)
[#permalink]
16 Nov 2021, 15:16
16 Nov 2021, 15:16
While this is true, there are two reasons I personally feel using derivatives is preferable to completing the square.
First, using derivatives to solve for a local minimum/maximum is easily generalizable whereas completing the square is not. For instance, completing the square is not helpful when evaluating a non-quadratic polynomial expression. On the other hand, in many advanced fields you use the derivative to compute a local minima/maxima, even for "complicated" functions.
Second, I personally find it a bit more straightforward since using the derivative to compute a local min/max is not a "trick," it's a well justified tool that makes sense and doesn't work just by chance.
I would like to add what is likely an unfounded belief, which is that I would think that anybody who is concerned enough about their quant GRE score to be looking at questions like this is applying to graduate school in a field that requires knowing calculus already. And in general, I would expect that most people are exposed to at least a minimal amount of calculus in college, and the notion of using the derivative/second derivative to solve for a local min/max is something you learn right at the start of any calculus course.
Ultimately, just because the GRE doesn't explicitly require calculus doesn't mean there aren't questions that are easier to solve with calculus. And in the case of this question, I personally think calculus is both the easiest and best justified tool to use (and not to overemphasize, but again this is just an opinion).
First, using derivatives to solve for a local minimum/maximum is easily generalizable whereas completing the square is not. For instance, completing the square is not helpful when evaluating a non-quadratic polynomial expression. On the other hand, in many advanced fields you use the derivative to compute a local minima/maxima, even for "complicated" functions.
Second, I personally find it a bit more straightforward since using the derivative to compute a local min/max is not a "trick," it's a well justified tool that makes sense and doesn't work just by chance.
I would like to add what is likely an unfounded belief, which is that I would think that anybody who is concerned enough about their quant GRE score to be looking at questions like this is applying to graduate school in a field that requires knowing calculus already. And in general, I would expect that most people are exposed to at least a minimal amount of calculus in college, and the notion of using the derivative/second derivative to solve for a local min/max is something you learn right at the start of any calculus course.
Ultimately, just because the GRE doesn't explicitly require calculus doesn't mean there aren't questions that are easier to solve with calculus. And in the case of this question, I personally think calculus is both the easiest and best justified tool to use (and not to overemphasize, but again this is just an opinion).
