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Grade 12th PassAlgebra

6)If x is real then the expression (x2 + 8x + 80)/(2x+8) cannot have any value between:
a)12 and 6 b)9 and -6 c)-6 and 9 d)8 and -8

Profile image of kshitij ahuja
14 Years agoGrade 12th Pass
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1 Answer

Profile image of Jitender Singh
11 Years ago
Ans:
Hello Student,
Please find answer to your question below

f(x) = \frac{x^{2}+8x+80}{2x+8}
For maxima & minima,
f'(x) = \frac{(2x+8)(2x+8) - 2(x^{2}+8x+80)}{(2x+8)^{2}}
f'(x) = \frac{(2x+8)^{2} - 2(x^{2}+8x+80)}{(2x+8)^{2}}
f'(x) = \frac{(4x^{2}+32x+64) - (2x^{2}+16x+160)}{(2x+8)^{2}}
f'(x) = \frac{2x^{2}+16x-96}{(2x+8)^{2}}
f'(x) = \frac{2(x^{2}+8x-48)}{(2x+8)^{2}}
f'(x) = \frac{2(x^{2}+8x-48)}{(2x+8)^{2}} = 0
\Rightarrow x = -12, 4
f(x) = \frac{x^{2}+8x+80}{2x+8}
f(4) = \frac{4^{2}+8.4+80}{2.4+8}
f(4) = \frac{128}{16} = 8
f(-12) = \frac{(-12)^{2}+8.(-12)+80}{2(-12)+8}
f(-12) = \frac{144-96+80}{-16}
f(-12) = -8
So f(x) have all values b/w -8 and 8.

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