What is the minimum value of the expression

Question

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}

Kinshook · Accepted Answer

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

IMO D

EgmatQuantExpert · Answer

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

https://e-gmat.com/blogs/wp-content/uploads/2018/07/FT-GC-FB-event-e1532524398949.jpg

TheNightKing · Answer

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

Thank you!

pk123 · Answer

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 02 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 Please give kudos if you like

TestPrepUnlimited · Answer

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.

SiffyB · Answer

Hi  TheNightKing ,

If you are aware of the concept to solve for local minima and maxima using differentiation, you can solve questions such as these like a breeze. I, too, have a hard time trying to figure out how to create an equation. Let me know if you need me to solve this question using differentation.

NOTE: I am fully aware of the fact that the GMAT doesn't test advanced concepts such as differentiation and integration, but IMO there's no harm employing them if they make your work easy!

tam87 · Answer

I generally always tend to use derivatives to find max/min value of an expression. However, in this case because of the fraction terms, won't to you think the calculations will be more using derivatives?

Posted from my mobile device

SiffyB · Answer

Hi  tam87 ,

You can rewrite the equation as:
 \frac{1}{x^2} –  \frac{1}{2x}  +  \frac{5}{4} 
=  x^-2 –  \frac{1}{2}x^-1  +  \frac{5}{4}

and then take the derivative from there: For example: derivative of  x^-2  = -2 ( x^-3 )

Hope that helps!

bumpbot · Answer

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What is the minimum value of the expression

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What is the minimum value of the expression

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