Finding range of $\frac{5}{(x+3)(x-4)}$

Question

Finding rage of $$\frac{5}{(x+3)(x-4)}$$

I first found the domain -

$x$ cannot be equal to $-3$ or $4$

I'm not too sure how do I go about to find the range , the way of thinking process to find the range . Thanks ..

Answer 1

Let $f(x)=\frac{5}{(x+3)(x-4)}$.

Hence, $f$ is a continuous function on $(-3,4)$, on $(-\infty,-3)$ and on $(4,+\infty)$.

Also, $$\lim_{x\rightarrow-3^-}f(x)=+\infty$$ and $$\lim_{x\rightarrow-\infty}f(x)=0,$$ which says that the range of $f$ on $(-\infty,-3)$ is $(0,+\infty)$.

By the same way we obtain that the range of $f$ on $(4,+\infty)$ is $(0,+\infty)$.

Now, $$\lim_{x\rightarrow-3^+}f(x)=-\infty,$$ $$\lim_{x\rightarrow4^-}f(x)=-\infty$$ and for $x\in(-3,4)$ by AM-GM we obtain $$f(x)=-\frac{5}{(x+3)(4-x)}\leq-\frac{5}{\left(\frac{3+x+4-x}{2}\right)^2}=-\frac{20}{49}.$$ The equality occurs for $x+3=4-x$,

which says that the range of $f$ on $(-3,4)$ is $$\left(-\infty,-\frac{20}{49}\right],$$ which says finally that tha range of $f$ is $$\mathbb R\setminus\left(-\frac{20}{49},0\right]$$.

Answer 2

Set $$y=\frac{5}{(x+3)(x-4)}$$ So we have that $$y(x+3)(x-4)=5.$$ We also know that the range of $y$ is the set of all numbers that make the discriminant greater than or equal to zero, in this case the discriminant is $$49y^2+20y.$$ The only values of $y$ where this is greater than or equal to zero is $y\leq -\frac{20}{49}$ and $y>0$, which is your range.

Answer 3

$$\dfrac5y=(x+3)(x-4)=x^2-x-12=\dfrac{(2x-1)^2-49}4$$

As for real $x,(2x-1)^2\ge0,$

$$\dfrac5y\ge-\dfrac{49}4$$

If $\dfrac5y=\dfrac{49}4, y=\dfrac{20}{49}$

Else $\dfrac5y>-\dfrac{49}4\iff\dfrac{49y+20}{4y}>0$

If $y>0, y>-\dfrac{20}{49}\implies y>\max\left(-\dfrac{20}{49},0\right)=?$

Else if $y<0,y<-\dfrac{20}{49}\implies y<\min\left(-\dfrac{20}{49},0\right)=?$

Answer 4

The domain is: $x\ne -3;4.$ The given rational function is continuously differentiable on its domain.

It must achieve its local max (min) value when the denominator ($x^2-x-12$) is min (max). (Alternatively, you can take derivative of the given function and equate it to $0$):

$$x_0=-\frac{-1}{2\cdot 1}=\frac12; \left(f'=-\frac{-5(2x-1)}{(x+3)^2(x-4)^2}=0 \Rightarrow x_0=\frac12.\right)$$ $$f \left(\frac12 \right)=-\frac{20}{49} \ \ \ \text{(local max)}.$$

Note: $$\lim_{x\to -3^-} f(x)=\lim_{x\to 4^+} f(x)=+\infty,$$ $$\lim_{x\to -3^+} f(x)=\lim_{x\to 4^-} f(x)=-\infty,$$ $$\lim_{x\to -\infty} f(x)=\lim_{x\to +\infty} f(x)=0.$$

Hence, the range is: $$y\in (-\infty, -\frac{20}{49}\bigg]\cup(0,+\infty).$$