What is the minimum value of f(x) = x^2 5x + 41?

Question

What is the minimum value of  f(x) = x^2 – 5x + 41 ?

A. 159/4
B. 149/4
C. 139/4
D. 129/4
E. 119/4

ScottTargetTestPrep · Accepted Answer

The function is a quadratic function of the form ax^2 + bx + c, and its graph is a parabola opening upward. The minimum value of the function is the y-value of the vertex of the parabola. Recall that the x-value of the vertex can be obtained by the formula: x = -b/(2a), so the x-value of the vertex is x = -(-5)/(2(1)) = 5/2. Therefore, the y-value of the vertex is y = f(5/2) = (5/2)^2 - 5(5/2) + 41 = 25/4 - 25/2 + 41 = 25/4 - 50/4 + 164/4 = 139/4.

Answer: C

SiffyB · Answer

I'll be using differentiation to solve this question:
 f(x) = x^2 – 5x + 41 
=  \frac{d f(x) }{dx}= \frac{d(x^2 – 5x + 41)}{dx} 
= 2x - 5

Let's equate this to 0 to get the minimum value of the function.
2x - 5 = 0

Thus, x = 5/2
Now, simply substitute this value of x in f(x)
 f(x) = x^2 – 5x + 41 
=  (5/2)^2 - 5(5/2) + 41 
=  25/4 - 25/2 + 41 
= 139/4

Answer (C)

GMATWhizTeam · Answer

Solution 
 •  f(x) = x^2 – 5x + 41 = x^2 – 2*\frac{5}{2}*x + \frac{25}{4} + \frac{139}{4} = (x – \frac{5}{2})^2 +\frac{139}{4} 
• Since, 139/4 is constant, f(x) will reach its minimum when  (x -\frac{5}{2})^2  is minimum.
 o And, minimum value of  (x -\frac{5}{2})^2 = 0  
• Thus, minimum value of  f(x) = 0 +\frac{139}{4} = \frac{139}{4}  
Thus, the correct answer is  Option C.

GMATinsight · Answer

f(x) = x^2 – 5x + 41

f(x) = x^2 – 2*(5/2)*x + (5/2)^2 + (41-25/4)

f(x) = (x – \frac{5}{2})^2 + (41-\frac{25}{4})

Now, for minimum value of this function  (x – \frac{5}{2})^2  must be zero

hence f (minimum) = 41-(25/4) = 139/4

Answer: Option C

itachiuchiha · Answer

For a quadratic equation  ax^2 + bx + c = 0 , the minimum value occurs at  \frac{-b}{2a}

In our equation, b = -5 and a = 1.

Minimum occurs at  \frac{-(-5)}{2 	imes 1} = \frac{5}{2}

The minimum value will be  ( \frac{5}{2} )^2 - 5 	imes \frac{5}{2} + 41

= \frac{25}{4} - \frac{25}{2} + 41

= \frac{-25}{4} + 41

= \frac{139}{4}

OA, C

BlueRocketAsh · Answer

f(x) = x^2 – 5x + 4 
 f(x) = (x-\frac{5}{2})^2+41-\frac{25}{4} 
 f(x) = (x-\frac{5}{2})^2+\frac{139}{4} 
Since it is a perfect square, minimum value of   (x-\frac{5}{2})^2  is 0
Therefore minimum value of the function is  \frac{139}{4}

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What is the minimum value of f(x) = x^2 5x + 41?

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