Last visit was: 12 Jul 2024, 16:43 It is currently 12 Jul 2024, 16:43

Close

GRE Prep Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GRE score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Close

Request Expert Reply

Confirm Cancel

Events & Promotions

Verbal Expert
Joined: 18 Apr 2015
Posts: 29080
Own Kudos [?]: 34063 [12]
Given Kudos: 25479
Send PM
Most Helpful Community Reply
avatar
Intern
Intern
Joined: 21 Jul 2019
Posts: 14
Own Kudos [?]: 32 [5]
Given Kudos: 0
Send PM
General Discussion
avatar
Intern
Intern
Joined: 27 Jun 2019
Posts: 40
Own Kudos [?]: 17 [1]
Given Kudos: 0
Send PM
Verbal Expert
Joined: 18 Apr 2015
Posts: 29080
Own Kudos [?]: 34063 [1]
Given Kudos: 25479
Send PM
Re: A function is defined by f(x) = x^2 + 4x – 5. What is the mi [#permalink]
Expert Reply
1
Bookmarks
Do not use calculus guys.

Use the straight way to have the correct answer
avatar
Intern
Intern
Joined: 11 Jul 2019
Posts: 33
Own Kudos [?]: 30 [1]
Given Kudos: 0
Send PM
Re: A function is defined by f(x) = x^2 + 4x – 5. What is the mi [#permalink]
1
what's the solution Mr.Carcass?
Verbal Expert
Joined: 18 Apr 2015
Posts: 29080
Own Kudos [?]: 34063 [0]
Given Kudos: 25479
Send PM
Re: A function is defined by f(x) = x^2 + 4x – 5. What is the mi [#permalink]
Expert Reply
shadowmr20 wrote:
what's the solution Mr.Carcass?



Those provided above by @jwolbrum are a perfect example of a flexible approach.

Nothing to add to those Sir
avatar
Intern
Intern
Joined: 14 Mar 2020
Posts: 41
Own Kudos [?]: 25 [0]
Given Kudos: 0
Send PM
Re: A function is defined by f(x) = x^2 + 4x – 5. What is the mi [#permalink]
jwolbrum wrote:
What's the source for this problem? I can see three relatively straightforward ways to solve it (verifying real roots, calculus, or the vertex formula for parabolas), but I haven't encountered questions that use these techniques on the GRE before.

I'm curious to see if anyone has an alternate solution.

Show: :: Plugging in and verifying that we have real roots
Start with the lowest value for f(x) and plug in.

f(x) = x^2 + 4x - 5
-9 = x^2 + 4x - 5

Set the quadratic equation equal to zero to factor:
0 = x^2 + 4x + 4
0 = (x+2)(x+2)
x = -2

Since we have non-imaginary roots, -9 is a reasonable value for f(x), and it was the smallest of our answer choices.


Show: :: Using Calculus
We can find the x-coordinate of the minimum or maximum of a function by taking the derivative and setting it equal to 0.

f(x) = x^2 + 4x - 5
f'(x) = 2x + 4
0 = 2x + 4
-4 = 2x
-2 = x

We can then plug in -2 as our x value and solve for f(x)

f(x) = (-2)^2 + 4(-2) - 5
f(x) = 4 - 8 - 5
f(x) = -9


Show: :: Using the parabola vertex formula
Given a parabola in the form f(x) = ax^2 +bx +c, we can identify the x-coordinate of the vertex using this formula:
vertex x-coordinate = -b / 2a

vertex x-coordinate = -4 / 2(1)
vertex x-coordinate = -2

We can then plug that x value into the equation and get the same result:

f(x) = (-2)^2 + 4(-2) - 5
f(x) = 4 - 8 - 5
f(x) = -9


while using ""Plugging in and verifying that we have real roots" method what the mean that we have non-imaginary roots?
Verbal Expert
Joined: 18 Apr 2015
Posts: 29080
Own Kudos [?]: 34063 [0]
Given Kudos: 25479
Send PM
Re: A function is defined by f(x) = x^2 + 4x – 5. What is the mi [#permalink]
1
Expert Reply
A function is defined by \(f(x) = x^2 + 4x – 5\).

What is the minimum value of \(f(x)\)?

\(f(x) = x^2 + 4x – 5 = (x+2)^2 -9\).
Since \((x+2)^2 =0\) for f(x) to be minimum
Minimum value of f(x) = -9

E
Moderator
Moderator
Joined: 02 Jan 2020
Status:GRE Quant Tutor
Posts: 1090
Own Kudos [?]: 903 [2]
Given Kudos: 9
Location: India
Concentration: General Management
Schools: XLRI Jamshedpur, India - Class of 2014
GMAT 1: 700 Q51 V31
GPA: 2.8
WE:Engineering (Computer Software)
Send PM
A function is defined by f(x) = x^2 + 4x – 5. What is the mi [#permalink]
1
1
Bookmarks
Given that \(f(x) = x^2 + 4x – 5\) and we need to find the minimum value of \(f(x)\)

\(f(x) = x^2 + 4x – 5\) = \(x^2 + 4x + 4 - 4 – 5\) = \(x^2 + 2*2x + 2^2 -9\) = \((x+2)^2 - 9\)

Now, we know that a Square of a number is always \(\geq\) 0
=> Minimum value of \((x+2)^2\) = 0
=> Minimum vale of f(x) = \((x+2)^2 - 9\) = 0-9 = -9

So, Answer will be E
Hope it helps!

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

Manager
Manager
Joined: 11 Jan 2022
Posts: 72
Own Kudos [?]: 11 [0]
Given Kudos: 437
Send PM
Re: A function is defined by f(x) = x^2 + 4x 5. What is the mi [#permalink]
What is the formula to find maximum?

Carcass wrote:
A function is defined by \(f(x) = x^2 + 4x – 5\).

What is the minimum value of \(f(x)\)?

\(f(x) = x^2 + 4x – 5 = (x+2)^2 -9\).
Since \((x+2)^2 =0\) for f(x) to be minimum
Minimum value of f(x) = -9

E
Intern
Intern
Joined: 26 Dec 2023
Posts: 44
Own Kudos [?]: 16 [0]
Given Kudos: 4
Send PM
Re: A function is defined by f(x) = x^2 + 4x 5. What is the mi [#permalink]
The way you solved it is much faster than my way, but I wanted to share my way in case it helped.

My instinct was to factor, so I did that, and found that it was (x+5)(x-1). Then, recognized it was a parabola, so the x-midpoint of the parabola's x-intercepts, is the x value when y is at its minimum.

The x-intercepts are x=-5 and x=1. Midpoint is -2. Plug in -2 into the equation and you get the answer, -9 (E).
GRE Instructor
Joined: 24 Dec 2018
Posts: 1036
Own Kudos [?]: 1379 [1]
Given Kudos: 24
Send PM
A function is defined by f(x) = x^2 + 4x 5. What is the mi [#permalink]
1
My instincts told me if we plug values less than zero, viz -1,-2,-3 we could get the answer quickly

If x = -1 f(x) = -8
If x = -2 f(x) = -9

We can stop at this point since -9 is the lowest value in the choices. The answer is Choice E.

If we proceed further

if x = -3 f(x) = -8

Note that f(x) decreases in value and increases once again (because it is a parabola). So it will never come back to -9 again. But we can try one more value:

if x = -4 f(x) = -5

So we can be sure -9 is the lowest value. This method required simple mental calculation and is probably faster than the other methods depending on your mental skills.
Moderator
Moderator
Joined: 02 Jan 2020
Status:GRE Quant Tutor
Posts: 1090
Own Kudos [?]: 903 [1]
Given Kudos: 9
Location: India
Concentration: General Management
Schools: XLRI Jamshedpur, India - Class of 2014
GMAT 1: 700 Q51 V31
GPA: 2.8
WE:Engineering (Computer Software)
Send PM
Re: A function is defined by f(x) = x^2 + 4x 5. What is the mi [#permalink]
1
Chaithraln2499: The maximum value will be infinity as the graph is a quadratic equation and as x increases to bigger values the value of the function will reach infinity.
Prep Club for GRE Bot
[#permalink]
Moderators:
GRE Instructor
46 posts
GRE Forum Moderator
17 posts
Moderator
1090 posts
GRE Instructor
218 posts

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne