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
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Sep 21Save up to 20% Target Test Prep GRE Plans
01:00 PM EDT
-11:59 PM PDT
Save up to $210 and get ready to crush your GRE score goals. Use the coupon code FLASH20 for the 1- and 6-month plans and coupon code FLASH15 for the LiveTeach bundle.
Sep 29How to Make your Deferred MBA Application Stand Out? | Deferred MBA Application
08:30 AM PDT
-09:30 AM PDT
we will be covering applications of some important deferred MBA programs such as HBS 2+2, ISB TLP, Yale Silver Scholar, ISB EEO, etc. Shruti Parashar, Founder of GoalISB Consulting, provides in-depth application strategies for the deferred MBA programs.
Sep 30End of Summer Discount!
03:00 PM EDT
-11:00 PM EDT
To end the summer, we are offering a one-time 10% discount on our Silver, Gold, and Platinum packages. If you are targeting R2, now is the time to buy! Valid for purchases made from 9/20 to 9/30.
Oct 03UNC Kenan-Flagler Virtual Open House
10:00 AM EDT
-11:00 AM EDT
Learn about the transformative power of an MBA from top-ranked UNC Kenan-Flagler. We will discuss the application process, curriculum and student life, plus anything else you're curious about. All attendees receive a $150 application fee waiver.
A function is defined by f(x) = x^2 + 4x – 5. What is the mi
[#permalink]
22 Jul 2019, 02:40
22 Jul 2019, 02:40
1
Expert Reply
9
Bookmarks
00:00
Question Stats:
45% (00:47) correct
54% (00:54) wrong based on 153 sessions
Hide Show timer Statistics
A function is defined by \(f(x) = x^2 + 4x – 5\). What is the minimum value of \(f(x)\)?
(A) 1
(B) 0
(C) –5
(D) –8
(E) –9
_________________
(A) 1
(B) 0
(C) –5
(D) –8
(E) –9
_________________
Re: A function is defined by f(x) = x^2 + 4x – 5. What is the mi
[#permalink]
22 Jul 2019, 10:47
22 Jul 2019, 10:47
5
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.
_________________
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.
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
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
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
_________________
-Jordan Wolbrum
JustMyWay Tutoring: personalized, individual tutoring for the LSAT, GMAT, and GRE.
jordan@justmywaytutoring.com
JustMyWay Tutoring: personalized, individual tutoring for the LSAT, GMAT, and GRE.
jordan@justmywaytutoring.com
General Discussion
Re: A function is defined by f(x) = x^2 + 4x – 5. What is the mi
[#permalink]
28 Aug 2019, 02:58
28 Aug 2019, 02:58
1
Bookmarks
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.
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.
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
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
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
I also took the derivative of the function and set it to zero.
Re: A function is defined by f(x) = x^2 + 4x – 5. What is the mi
[#permalink]
28 Aug 2019, 10:25
28 Aug 2019, 10:25
Expert Reply
1
Bookmarks
Do not use calculus guys.
Use the straight way to have the correct answer
_________________
Use the straight way to have the correct answer
_________________
Re: A function is defined by f(x) = x^2 + 4x – 5. What is the mi
[#permalink]
29 Aug 2019, 01:39
29 Aug 2019, 01:39
1
what's the solution Mr.Carcass?
Re: A function is defined by f(x) = x^2 + 4x – 5. What is the mi
[#permalink]
29 Aug 2019, 07:10
29 Aug 2019, 07:10
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
_________________
Re: A function is defined by f(x) = x^2 + 4x – 5. What is the mi
[#permalink]
10 Apr 2020, 17:45
10 Apr 2020, 17:45
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.
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.
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
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
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?
Re: A function is defined by f(x) = x^2 + 4x – 5. What is the mi
[#permalink]
11 Apr 2020, 03:36
11 Apr 2020, 03:36
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
_________________
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
Joined: 02 Jan 2020
Status:GRE Quant Tutor
Posts: 1040
Location: India
Concentration: General Management
Schools: XLRI Jamshedpur, India - Class of 2014
GMAT 1: 700 Q51 V31
GPA: 2.8
WE:Engineering (Computer Software)
A function is defined by f(x) = x^2 + 4x – 5. What is the mi
[#permalink]
08 Jan 2022, 01:44
08 Jan 2022, 01:44
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
_________________
\(f(x) = x^2 + 4x – 5\) = \(x^2 + 4x + 4 - 4 – 5\) = \(x^2 + 2*2x + 2^2 -9\) = \((x+2)^2 - 9\)
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
_________________
Ankit
to BrushMyQuant YouTube Channel to get WEEKLY new VIDEOS!!!
How to Solve :
Equation of a Line || Co-ordinate Geometry || Remainders || LCM and GCD || Inequalities || Absolute Value || Factors and Multiples || Number and Sum of Factors || Similar Triangles || Exponents || Work Rate || Probability: Coin Toss || Probability: Dice Roll || Prime Numbers || Circles || Equilateral Triangle in a Circle || Arithmetic and Geometric Progression || Divisibility Rules || Statistics || Functions and Custom Characters || Stem & Leaf Plot || Box And Whisker Plot
GRE Club Topic Wise Problems
GRE Quant - 2019 Ed
GRE Verbal - 2019 Ed
The Princeton Review Quantitative Directory
The 5 lb. Book of GRE Practice Problems (2nd Ed.) - QUANT
GRE - PowerPrep® The FREE Practice Tests Explained 2019
to BrushMyQuant YouTube Channel to get WEEKLY new VIDEOS!!!How to Solve :
Equation of a Line || Co-ordinate Geometry || Remainders || LCM and GCD || Inequalities || Absolute Value || Factors and Multiples || Number and Sum of Factors || Similar Triangles || Exponents || Work Rate || Probability: Coin Toss || Probability: Dice Roll || Prime Numbers || Circles || Equilateral Triangle in a Circle || Arithmetic and Geometric Progression || Divisibility Rules || Statistics || Functions and Custom Characters || Stem & Leaf Plot || Box And Whisker Plot
GRE Club Topic Wise Problems
GRE Quant - 2019 Ed
GRE Verbal - 2019 Ed
The Princeton Review Quantitative Directory
The 5 lb. Book of GRE Practice Problems (2nd Ed.) - QUANT
GRE - PowerPrep® The FREE Practice Tests Explained 2019
Re: A function is defined by f(x) = x^2 + 4x 5. What is the mi
[#permalink]
03 May 2022, 02:35
03 May 2022, 02:35
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
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
Re: A function is defined by f(x) = x^2 + 4x 5. What is the mi
[#permalink]
15 Jun 2023, 22:50
15 Jun 2023, 22:50
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.
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.
gmatclubot
Moderators:
|
|
||
Hi Guest,Here are updates for you: |
B