Answer:
Domain : (- ∞, - 5), and (3 / 2, ∞),
Range : (∞, 0.4437]
Explanation:
Assuming that we want our answer in interval notation, let's start by determining the domain. Remember that the domain can be found where the function is undefined.
Given : f(x) = (2x - 7) / (2x² + 7x - 15)
Alternative Form : (2x - 7) / (x + 5)(2x - 3)
To receive this 'alternative form' we can simply factor the expression 2x² + 7x - 15. See the procedure below,
![\mathrm{Given : 2x^2+7x-15} ,](https://img.qammunity.org/2021/formulas/mathematics/college/pd46vy0dlvnlhvywnfgn4rw49b5qtc5pqk.png)
![\mathrm{Break\:the\:expression\:into\:groups : \left(2x^2-3x\right)+\left(10x-15\right)} ,](https://img.qammunity.org/2021/formulas/mathematics/college/2ia960bs96wtzmfequts52n6pvilke0w3m.png)
![\mathrm{Factor\:out\:}x\mathrm{\:from\:}2x^2-3x\mathrm{:\quad }x\left(2x-3\right),\\\mathrm{Factor\:out\:}5\mathrm{\:from\:}10x-15\mathrm{:\quad }5\left(2x-3\right),](https://img.qammunity.org/2021/formulas/mathematics/college/lnyyarh0rcq59jdhkyhs4ksmha6vcln023.png)
![x\left(2x-3\right)+5\left(2x-3\right) = \left(2x-3\right)\left(x+5\right) - \mathrm{Factored\:expression}](https://img.qammunity.org/2021/formulas/mathematics/college/f42euuod8cx8hu9m5yq126vjzn8myrntkn.png)
Now let's find the domain using the expression '(x + 5)(2x - 3) = 0.' If the denominator equals 0, the function is considered undefined.
![\mathrm{Given : \left(x+5\right)\left(2x-3\right)=0} ,\\x+5=0:\quad x=-5, 2x-3=0:\quad x=(3)/(2)\\\mathrm{The\:solutions\:are: x=-5,\:x=(3)/(2)}](https://img.qammunity.org/2021/formulas/mathematics/college/rhqsns7zlfnukleda76o3gz88mj2fppwa3.png)
Knowing these solutions the domain has the intervals (- ∞, - 5), and (3 / 2, ∞). The range is the set of values that correspond to the domain, so in this case the range would be (∞, 42 + 8√17 / 169]. 42 + 8√17 / 169 = (About) 0.4437, so it lies on the interval (∞, 0.4437].
The images below represent the plotted function in two areas. There are 3 curves in this graph.