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What is the minimal value of f(n) = 2n^2 - 7n + 8, if n is an integer?

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What is the minimal value of f(n) = 2n^2 - 7n + 8, if n is an integer?

by Max@Math Revolution » Thu Mar 19, 2020 1:51 am

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[GMAT math practice question]

What is the minimal value of f(n) = 2n^2 - 7n + 8, if n is an integer?

A. 1
B. 2
C. 3
D. 4
E. 5
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Re: What is the minimal value of f(n) = 2n^2 - 7n + 8, if n is an integer?

by deloitte247 » Sat Mar 21, 2020 2:46 pm
Given that f(n) = 2n^2 - 7n + 8
The given function is quadratic of form ax^2 + bx + c.
The minimum value of 'n' can be gotten with the equation for finding the vertex of a quadratic function = -b/2a, where b=-7, and a=2.
$$-\frac{\left(-7\right)}{2\cdot2}=\frac{7}{4}=1.75$$
The minimum value of n=1.75, but since n is an integer, the nearest function to 1.75 = 2.
So, for the function f(n) = 2n^2 - 7n + 8
when n=2
f(2) = 2(2)^2 - 7(2) + 8
= 2(4) - 14 + 8
= 8 - 14 + 8
= 2

Answer = B
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Re: What is the minimal value of f(n) = 2n^2 - 7n + 8, if n is an integer?

by Max@Math Revolution » Sun Mar 22, 2020 6:52 pm
=>

f(n) = 2n^2-7n+8
= 2(n^2 - (7/2)n) + 8 (taking out a common factor of 2 from the first 2 terms)
= 2(n^2 - 2(7/4)n + (49/16) - (49/16)) + 8 (completing the square)
= 2(n - 7/4)^2 - 49/8 + 8 (factoring and multiplying 2 times 49/8)
= 2(n - 7/4)^2 + 15/8 (adding like terms)
f(n) has a minimum value when n = 7/4
The closest integer to 7/4 is 2.
Thus the minimum value of f(n) is f(2) = 2*2^2 - 7*2 + 8 = 8 – 14 + 8 = 2.

Therefore, B is the answer.
Answer: B
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