Asked by Athenal
Someone help me!!
square root = sqrt
(5x^2-5x-10)sqrt(x+7) + (2x+6)sqrt(x+2) >= x^3+13x^2-6x-28
thank you!
square root = sqrt
(5x^2-5x-10)sqrt(x+7) + (2x+6)sqrt(x+2) >= x^3+13x^2-6x-28
thank you!
Answers
Answered by
Steve
well, looking at the graphs, it looks like -2 <= x <= 2
checking, if we call them
f(x)+g(x) >= h(x), then
f(-2) = 20√5
g(-2) = 0
f(2) = 0
g(2) = 20
h(-2) = 28
h(2) = 20
so,
f(-2)+g(-2) >= h(-2)
f(2) + g(2) >= h(2)
Clearly we need x >= -2 for √(x+2) to be real.
Beyond x=2, h(x) clearly grows faster than f+g, since h is a cubic (x^3) and f+g is basically x^2.5, so for x>2, f+g < h
The graph of (f+g)-h is shown here:
http://www.wolframalpha.com/input/?i=plot+y%3D%285x^2-5x-10%29%E2%88%9A%28x%2B7%29+%2B+%282x%2B6%29%E2%88%9A%28x%2B2%29+,+y%3Dx^3%2B13x^2-6x-28+for+-2+%3C%3D+x+%3C%3D+3
checking, if we call them
f(x)+g(x) >= h(x), then
f(-2) = 20√5
g(-2) = 0
f(2) = 0
g(2) = 20
h(-2) = 28
h(2) = 20
so,
f(-2)+g(-2) >= h(-2)
f(2) + g(2) >= h(2)
Clearly we need x >= -2 for √(x+2) to be real.
Beyond x=2, h(x) clearly grows faster than f+g, since h is a cubic (x^3) and f+g is basically x^2.5, so for x>2, f+g < h
The graph of (f+g)-h is shown here:
http://www.wolframalpha.com/input/?i=plot+y%3D%285x^2-5x-10%29%E2%88%9A%28x%2B7%29+%2B+%282x%2B6%29%E2%88%9A%28x%2B2%29+,+y%3Dx^3%2B13x^2-6x-28+for+-2+%3C%3D+x+%3C%3D+3
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