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If f(2x - 1) = 4x² - 4x - 8, then f(5) =

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GreenlightTestPrep
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If f(2x - 1) = 4x² - 4x - 8, then f(5) = [#permalink] New post   17 Apr 2019, 09:50
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If f(2x - 1) = 4x² - 4x - 8, then f(5) =

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GreenlightTestPrep
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Re: If f(2x - 1) = 4x² - 4x - 8, then f(5) = [#permalink] New post   19 Apr 2019, 04:35
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GreenlightTestPrep wrote:
If f(2x - 1) = 4x² - 4x - 8, then f(5) =

Show: ::
16


Approach #1:
Solve the equation 2x - 1 = 5 to get x = 3

Take: f(2x - 1) = 4x² - 4x - 8
Plug in x = 3 to get: f(2(3) - 1) = 4(3)² - 4(3) - 8
Simplify: f(5) = 36 - 12 - 8 = 16

Answer: 16

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Brent
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Re: If f(2x - 1) = 4x² - 4x - 8, then f(5) = [#permalink] New post   19 Apr 2019, 04:36
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GreenlightTestPrep wrote:
If f(2x - 1) = 4x² - 4x - 8, then f(5) =

Show: ::
16


Approach #2:
First notice that 4x² - 4x - 8 looks a lot like (2x - 1)²

(2x - 1)² = (2x - 1)(2x - 1) = 4x² - 4x + 1

Now take: f(2x - 1) = 4x² - 4x - 8
Rewrite as: f(2x - 1) = 4x² - 4x + 1 - 9 = (2x - 1)² - 9
In other words, f(something) =something² - 9

In general, we can write f(x) = x² - 9

We're asked to evaluate f(5)
So f(5) = 5² - 9 = 25 - 9 = 16

Answer: 16

Cheers,
Brent
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HarishKumar
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If f(2x - 1) = 4x² - 4x - 8, then f(5) = [#permalink] New post   14 Feb 2024, 20:59
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Look for patterns as GRE is not a test of your calculation abilities.

(2x−1)2=4x2−4x+1

Therefore,

f(2x−1)=4x2−4x−8 can be written as

f(2x−1)=(2x−1)2−9

to compensate for the extra constant 1 produced by (2x−1)2 along with 4x2−4x.

Now,

f(5)=52−9=25−9=16

The answer is 16
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If f(2x - 1) = 4x² - 4x - 8, then f(5) = [#permalink]
14 Feb 2024, 20:59
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