Page 106 - Elementary Algebra Exercise Book I

ELEMENTARY ALGEBRA EXERCISE BOOK I inequAlities

3.57 Given a< −1, and x satisfies x + ax ≤−x , and x + ax has the minimum
2
2
1
value − , find the value of a .
2
2 2 a 2 a 2
Solution: a< −1,x + ax ≤−x ⇒ x[x +(a + 1)] ≤ 0 ⇒ 0 ≤ x ≤−(a + 1). Let f(x) = x + ax =(x + ) − 4

2
2 a 2 a 2
f(x) = x + ax =(x + ) − .
2 4
If −(a + 1) < − ⇔−2 <a < −1, then f(x) reaches its minimum value f(−a − 1) = a +1
a
2
3
1
at x = −(a + 1), thus a +1 = − ⇒ a = − .
2 2
2
a
If −(a + 1) ≥− ⇔ a ≤−2, then f(x) reaches it minimum value − at x = − , thus
a
a
2 4 2
√
a 2 1
− = − ⇒ a = ± 2 both of which violate a ≤−2.
4 2
As a conclusion, a = − .
3
2

This e-book

is made with SETASIGN

SetaPDF

PDF components for PHP developers

www.setasign.com

Download free eBooks at bookboon.com Click on the ad to read more
Click on the ad to read more
Click on the ad to read more
Click on the ad to read more
Click on the ad to read more
Click on the ad to read more
Click on the ad to read more
Click on the ad to read more
Click on the ad to read more
Click on the ad to read more
Click on the ad to read more
Click on the ad to read more
Click on the ad to read more
Click on the ad to read more
Click on the ad to read more
Click on the ad to read more
Click on the ad to read more
Click on the ad to read more
Click on the ad to read more
Click on the ad to read more
Click on the ad to read more
Click on the ad to read more
Click on the ad to read more
Click on the ad to read more
Click on the ad to read more
Click on the ad to read more
Click on the ad to read more
Click on the ad to read more
Click on the ad to read more
Click on the ad to read more
Click on the ad to read more
Click on the ad to read more
Click on the ad to read more
Click on the ad to read more
106
106