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What is the minimum value of the expression

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What is the minimum value of the expression  [#permalink]

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New post 31 Jul 2019, 05:17
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A
B
C
D
E

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  95% (hard)

Question Stats:

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Algerba Question Series- Question 3




What is the minimum value of the expression \(\frac{1}{x^2}\)– \(\frac{1}{2x}\) + \(\frac{5}{4}\)?

    A. \(\frac{-5}{4}\)
    B. \(\frac{-19}{16}\)
    C. 0
    D. \(\frac{19}{16}\)
    E. \(\frac{5}{4}\)

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What is the minimum value of the expression  [#permalink]

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New post 31 Jul 2019, 08:41
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EgmatQuantExpert wrote:

Algerba Question Series- Question 3




What is the minimum value of the expression \(\frac{1}{x^2}\)– \(\frac{1}{2x}\) + \(\frac{5}{4}\)?

    A. \(\frac{-5}{4}\)
    B. \(\frac{-19}{16}\)
    C. 0
    D. \(\frac{19}{16}\)
    E. \(\frac{5}{4}\)




What is the minimum value of the expression \(\frac{1}{x^2}– \frac{1}{2x} + \frac{5}{4}\)
\(t^2– \frac{t}{2} + \frac{5}{4}\) where \(t=\frac{1}{x}\)
\(t^2 - \frac{2t}{4} + \frac{1}{16} +\frac{5}{4}-\frac{1}{16}\)
\(=(t-\frac{1}{4})^2 + \frac{19}{16}\)
=\frac{19}{16} since minimum value of expression is when \((t-\frac{1}{4})^2 =0\)

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Re: What is the minimum value of the expression  [#permalink]

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New post 05 Aug 2019, 22:01

Solution


Given:
    • An expression \(\frac{1}{x^2} – \frac{1}{2x} + \frac{5}{4}\)

To find:
    • The minimum value of the expression \(\frac{1}{x^2} – \frac{1}{2x} + \frac{5}{4}\)

Approach and Working Out:
    • \(\frac{1}{x^2} – \frac{1}{2x} + \frac{5}{4} = (\frac{1}{x})^2 – 2 * \frac{1}{x} * \frac{1}{4} + (\frac{1}{4})^2 + \frac{19}{16} = (\frac{1}{x} – \frac{1}{4})^2 + \frac{19}{16}\)

Therefore, the minimum value of the expression \(\frac{1}{x^2} – \frac{1}{2x} + \frac{5}{4} = 0 + \frac{19}{16} = \frac{19}{16}\)

Hence, the correct answer is Option D.

Answer: D

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Re: What is the minimum value of the expression  [#permalink]

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New post 23 Oct 2019, 09:08
Any other approach apart from re-writing the equation in the perfect square standard form? It is least one of favorites.

Thank you!
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What is the minimum value of the expression  [#permalink]

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New post 23 Oct 2019, 10:14
TheNightKing wrote:
Any other approach apart from re-writing the equation in the perfect square standard form? It is least one of favorites.

Thank you!



What is the minimum value of the expression 1/X^2 -1/2x +5/4

Lets take x <0, in this case 1/x^2 is +ve & -1/2x also becomes +ve

Now if x>0, for any value of x, expression 1/x^2 -1/2x will be +ve if 0<x<2...how lets take x=0.5 1/x^2= 1/0.25=4 while -1/2x= -1 so the entire expression is +ve
but when x>2 lets take x= 4, 1/4^2 = 1/16=0.0625 while -1/2x = -1/8=-0.125 hence expression becomes -ve
lets take x a large expression this entire (1/x^2 -1/2x) will not be smaller than -5/4

hence entire expression will always be more than 0
Now we are left with 2 options D or E
E is possible only when x is infinite and for any value less than infinite but as discussed for x>2 , the expression 1/x^2 -1/2x can be <0 which will make entire expression of 1/x^2-1/2x+5/4 slightly less than 5/4 which is only option D

Not sure the above helps

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What is the minimum value of the expression  [#permalink]

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New post 23 Oct 2019, 12:12
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We can look at the answers to give ourselves a hint. The denominator for the combined expression is dominated by \(x^2\), as that gives the biggest denominator among all fractions. Starting from A and E, if we want a denominator of 4 we plug in x = 2 or x = -2. x = -2 gives a denominator of 4, however the coefficient for \(\frac{-1}{2x}\) is negative so a positive x gives a smaller value. Then x = 2 gives 1/4 - 1/4 + 5/4 = 5/4 which is a good start.
Next, looking at choices B and D, we need to try to get a denominator of 16, which requires x = 4. Plugging that in gives 1/16 - 1/8 + 5/4 = 19/16. This is smaller than 5/4 so 19/16 is the new minimum. Finally, there are no more denominators to test thus we choose D.

EgmatQuantExpert wrote:

Algerba Question Series- Question 3




What is the minimum value of the expression \(\frac{1}{x^2}\)– \(\frac{1}{2x}\) + \(\frac{5}{4}\)?

    A. \(\frac{-5}{4}\)
    B. \(\frac{-19}{16}\)
    C. 0
    D. \(\frac{19}{16}\)
    E. \(\frac{5}{4}\)


TheNightKing wrote:
Any other approach apart from re-writing the equation in the perfect square standard form? It is least one of favorites.

Thank you!

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What is the minimum value of the expression   [#permalink] 23 Oct 2019, 12:12
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