Show $2 \lt f(x) \lt 2.8$?

Question

I have a question about a function $f(x)$: $$f(x) = -x + \frac{x^2}{x-2} - \frac{20}{x^2 + x - 6},\qquad x>2$$ simplifies to $$f(x) = 2x + \frac{10}{x + 3},\qquad x>2$$ How would you show the range of $f(x)$ is $2 < f(x) < 2.8?$ Thank you.

I think it should be $$\frac{2x+10}{x+3}$$ – lab bhattacharjee May 12 '14 at 13:50 — lab bhattacharjee, May 12 '14 at 13:50

lab bhattacharjee · Answer 1 · 2014-05-12T13:52:31.847

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I think it should be $$\frac{2x+10}{x+3}=\frac{2(x+3)+4}{x+3}=2+\frac4{x+3}$$

As $\displaystyle x>2,x+3>5\implies0<\frac1{x+3}<\frac15$

edited May 12 '14 at 13:52

answered May 12 '14 at 13:45

lab bhattacharjee

274,582

based on what I calculated, he is correct. – May 12 '14 at 13:47
@jonnytan999, Edited – lab bhattacharjee May 12 '14 at 13:53
ah! I was halfway through typing the answer when this came... – May 12 '14 at 14:03
im not quite sure what you did with the (2x+10)/(x+3) ? – Yusuf May 12 '14 at 14:16
ohh sorry, i see so is the range of the function what your answer is? – Yusuf May 12 '14 at 14:18
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@user149550, if $0<\frac1{x+3}<\frac15,$ can't you find the range of $f(x)?$ – lab bhattacharjee May 12 '14 at 14:45
@lab bhattacharjee im not really sure how to? – Yusuf May 12 '14 at 15:07
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@user149550, multiply each side of $$0<\frac1{x+3}<\frac15$$ by $4$. Then add $2$ to each side – lab bhattacharjee May 12 '14 at 15:13
@lab bhattacharjee why would you do that? – Yusuf May 12 '14 at 15:16

Show $2 \lt f(x) \lt 2.8$?

1 Answers1