What would be the range for the function $f(x)=\log_3(5+4x-x^2)$?

Question

I need to find the range of the function $$f(x)=\log_3(5+4x-x^2)$$

Here's what I tried to do:

First I found the domain, by setting, $$5+4x-x^2 > 0$$

$$\implies x^2-5-4x<0$$ $$\implies (x-5)(x+1)<0$$ So, from here, the domain is $x \in (-1,5)$

So here the domain is a finite open interval. Also, I know that, $\log_ak$, where $a>1$ is an increasing function. But I have no idea where to go from here. A detailed explanation would be much appreciated.

Answer 1

$\frac d {dx} (5+4x-x^{2})=4-2x$. This is positive in $(-1,2)$ and negative in $(2,5)$. So the function $5+4x-x^{2}$ is increasing up to $x=2$ and then it is decreasing. Its maximum is $9$ and its infimum is $0$ (the limit of he function as $ x\to -1$ or $x \to 5$). Hence the maximum value of $f$ is $\log_3 9=2$ and there is no minimum. [$\log_3 x \to -\infty$ as $x$ tends to $0$ through positive values].

Answer 2

We have that $5+4x-x^2 = -(x-2)^2 +9.$ The range of this function is $(-\infty, 9],$ since logarithm is defined for positive numbers only, we consider the interval $(0, 9].$ Since logarithm is increasing and continuous function, the range of $f$ is $(-\infty, 2].$

Edit:

Again consider the graph of $g(x)= \log_3(x).$ Look at the graph carefully. Our function $g$ is NOT defined for negative numbers. Furthermore, if $x$ approaches to $0,$ then $\log_3(x)$ approaches to $-\infty.$ Does this help?