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Re: A function is defined by f(x) = x^2 + 4x – 5. What is the mi [#permalink]
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Do not use calculus guys.

Use the straight way to have the correct answer
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Re: A function is defined by f(x) = x^2 + 4x – 5. What is the mi [#permalink]
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what's the solution Mr.Carcass?
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Re: A function is defined by f(x) = x^2 + 4x – 5. What is the mi [#permalink]
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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
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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?
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Re: A function is defined by f(x) = x^2 + 4x – 5. What is the mi [#permalink]
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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
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A function is defined by f(x) = x^2 + 4x – 5. What is the mi [#permalink]
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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

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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
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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).
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A function is defined by f(x) = x^2 + 4x 5. What is the mi [#permalink]
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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.
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Re: A function is defined by f(x) = x^2 + 4x 5. What is the mi [#permalink]
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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.
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Re: A function is defined by f(x) = x^2 + 4x 5. What is the mi [#permalink]
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