# What is the range of the function, $f(x)=\dfrac{4x+7}{(6x)^2-5}$, algebraically?

What is the range of the function, $f(x)=\dfrac{4x+7}{(6x)^2-5}$, algebraically?

I have a problem which I am stuck in. I know that the range is equivalent to the domain of the inverse function. I have tried to find the inverse the function but got confuse when isolating y.

• You need to differentiate it to find minima and maxima and find limits in infinities and every point the function is not continuous. Apr 26, 2018 at 4:57
• Your function is not one-one. You can't find an inverse without restricting the domain, which will further complicate the matter. Apr 26, 2018 at 5:15

Consider the intersection between $y=k$ and $y=f(x)$, we have$$k=\dfrac{4x+7}{(6x)^2-5}$$ Rearranging $$(36k)x^2-4x-(7+5k)=0$$ For intersection to occur, $$(-4)^2-4(36k)(-1)(7+5k)\ge0$$ Simplifying, $$45k^2+63k+1\ge0$$ Completing the square, $$\frac{900}{421}\left(k+\frac7{10}\right)^2-1\ge0$$ So $$k\ge\frac1{30}\left(\sqrt{421}-21\right),\quad k\le-\frac1{30}\left(\sqrt{421}+21\right)$$
Hence, the range of $f(x)$ is $$\left(-\infty,-\frac1{30}\left(\sqrt{421}+21\right)\right]\cup\left[\frac1{30}\left(\sqrt{421}-21\right),\infty\right)$$
• @Soroushkhoubyarian $$\sqrt{421}<21$$ Hence, $$\frac1{30}\left(\sqrt{421}-21\right)<0$$ Apr 26, 2018 at 5:17
You can show first that $f'(x) < 0$, hence $f$ is one-to-one and has an inverse. Now you can find it. Write $y = \dfrac{4x+7}{36x^2-5}$,and switch $x$ and $y$ to get: $x = \dfrac{4y+7}{36y^2-5}$. Now solve for $y$. We have: $36xy^2-4y-5x-7 = 0\implies y = \dfrac{-(-4)\pm \sqrt{(-4)^2-4(36x)(-5x-7)}}{2(36x)}= \dfrac{4 \pm \sqrt{720x^2+1008x + 16}}{72x}= f^{-1}(x)$. You can now find the domain of $f^{-1}(x)$ and call it $D_{f^{-1}}$, then $D_{f^{-1}}= A \cap B$, whereas $A = \mathbb{R}\setminus\{0\}, B = \{x: 720x^2+1008x+16 \ge 0\}$. And you would get the same answer as the first post.