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- Thread starter Vali
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- Apr 22, 2018

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I presume you got

$$f'(x)\ =\ \frac{2(x+3)}{(x-1)^3}$$

which is correct. Notice that $f'(x)<0$ when $x\in[0,\,1)$ and $f'(x)>0$ when $x\in(1,\,3]$. So $f(x)$ is decreasing on $[0,\,1)$ and increasing on $(1,\,3]$. Also, $f(0)=f(3)=-2$ and $f(x)\to-\infty$ as $x$ tends to $1$ from either side. From these, you should be able to piece together the range of $f(x)$.

$$f'(x)\ =\ \frac{2(x+3)}{(x-1)^3}$$

which is correct. Notice that $f'(x)<0$ when $x\in[0,\,1)$ and $f'(x)>0$ when $x\in(1,\,3]$. So $f(x)$ is decreasing on $[0,\,1)$ and increasing on $(1,\,3]$. Also, $f(0)=f(3)=-2$ and $f(x)\to-\infty$ as $x$ tends to $1$ from either side. From these, you should be able to piece together the range of $f(x)$.

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