If $f(x)=\frac{1}{x}-\frac{4}{\sqrt{x}}+3$ where $\frac{1}{16}\le x\le 1$, find the range of $f(x)$. $ \textbf{(A) }-2\le f(x)\le 4 \qquad\textbf{(B) }-1\le f(x)\le 3\qquad\textbf{(C) }0\le f(x)\le 3\qquad\textbf{(D) }-1\le f(x)\le 4\qquad\textbf{(E) }\text{None of the above} $
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Tags: function
roza2010
03.06.2014 10:02
Function $f(x)=\frac{1}{x}-\frac{4}{\sqrt{x}}+3$ is decrease in $[1/16,1/4)$ and increase in $(1/4,1]$ , so range is $[f(1/4),f(1)]$ , or $[-1,0]$ , so (E) is true.
SMOJ
03.06.2014 10:24
What if $x=\frac{1}{16}$?
roza2010
04.06.2014 02:17
local minima
HapaxOromenon
07.05.2018 21:16
roza2010 wrote: local minima The answer is B, because x = 1/16 gives the maximum value of 3.