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Q. If `x` is real,then least value of expression `(x^2-6x+5)/(x^2+2x+1)` is: A) `-1`...

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user8235304 | Student, Grade 11 | (Level 1) Valedictorian

Posted August 16, 2013 at 5:26 PM via web

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Q. If `x` is real,then least value of expression `(x^2-6x+5)/(x^2+2x+1)` is:

A) `-1`

B) `-(1/2)`

C) `-(1/3)`

D) none of these

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted August 16, 2013 at 6:06 PM (Answer #1)

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The least value of (x^2 - 6x + 5)/(x^2 + 2x + 1) has to be determined if x is real.

If f(x) = (x^2 - 6x + 5)/(x^2 + 2x + 1), the least value of f(x) is f(a) where a is the solution of f'(x) = 0 and f''(a) is positive.

f'(x) = (8*x-16)/(x^3+3*x^2+3*x+1)

f'(x) = 0

=> 8x - 16 = 0

=> x = 2

f''(x) = -(16*x-56)/(x^4+4*x^3+6*x^2+4*x+1)

f''(2) `~~` 0.296296 which is positive.

For x = 2, (x^2 - 6x + 5)/(x^2 + 2x + 1) = (4 - 12 + 5)/(4 + 4 + 1) = -3/9 = -1/3

The least value of the given expression is -1/3.

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