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Suppose there is a function $$f(x)=\log_3(5+4x-x^2)$$ Then find the range of $f(x)$.

I don't know how to find range of functions like $g(x)=\log_n(h(x))$ But still I decided to give it a try.

Since $f(x)$ is a log function, we have $x\in(-1,5)$ Let us denote $5+4x-x^2=j(x)$

So the minimum value of $j(x)$ in the interval $x\in(-1,5)$ is $\lim j(x)\rightarrow0$ and the maximum value is $\lim j(x)\rightarrow5$. After this I am stuck.

I actually want to know how to find range of a function like this $$b(x)=\log_n(ax^2+bx+c)$$

Any help is greatly appreciated.

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3 Answers 3

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$5+4x-x^2=-(x^2-4x+4)+9=-(x-2)^2+9=(3-(x-2))(3+(x-2))=-(x+1)(x-5)$

Sine $-(x+1)(x-5)>0$ when $x\in(1,5)$, the function is defined when $x\in(1,5)$.

On this interval $f(x)\to-\infty$ as $x\to -1$ and $f(x)\to-\infty$ as $x\to 5$.

Next, $f'(x)=\frac{4-2x}{\ln(3)(5+4x-x^2)}$.

$f'(x)=0$, when $x=2$.

At this point there is a maximum, thus the range of the function is $(-\infty, f(2)]=(-\infty, \log_3(9)]$.

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Logarithm to any base is a strictly increasing continuous function. The range of any continuous fucntion on $\mathbb R$ or an interval in $\mathbb R$ is an interval and so the range of $\log_n (f(x))$ is simply the interval from $f(a)$ to $f(b)$ where $a,b$ are the end points of the arnge of $f$. I am ignoring the end points here but you should be able to see whether the end poinst are included or not in the present case. [The answer in this case is $(-\infty, 2]$].

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There is no minimum for $f$. There is a maximum somewhere, and you need to find it. Say the maximal value is $m$, then the range is $(-\infty,m]$. In fact, it is easy to see that the maximum is attained at $x=2$, so that the maximal value is $m=\log_3(9)=2$, hence the range is $(-\infty,2]$.

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