Question

what is the domain of the rational function ( f(x) = \frac{2x}{2x^3 + 5x^2 - 12x} )?

• ( {x in mathbb{r} mid x

eq 0} )

• ( {x in mathbb{r} mid x

eq -\frac{3}{2}, 4} )

• ( {x in mathbb{r} mid x

eq 0, -\frac{3}{2}, 4} )

• ( {x in mathbb{r} mid x

eq 0, \frac{3}{2}, -4} )

Explanation:

Step1: Factor the denominator

First, factor the denominator \(2x^{3}+5x^{2}-12x\). We can factor out an \(x\) first: \(x(2x^{2}+5x - 12)\). Then, factor the quadratic \(2x^{2}+5x - 12\). We need two numbers that multiply to \(2\times(-12)=-24\) and add to \(5\). Those numbers are \(8\) and \(-3\). So, \(2x^{2}+5x - 12 = 2x^{2}+8x - 3x - 12=2x(x + 4)-3(x + 4)=(2x - 3)(x + 4)\). So the denominator factors to \(x(2x - 3)(x + 4)\) or \(x(2x - 3)(x + 4)=2x^{3}+5x^{2}-12x\) (we can also factor out a 2 from the first factoring step? Wait, no, original denominator is \(2x^{3}+5x^{2}-12x\), so factoring out \(x\) gives \(x(2x^{2}+5x - 12)\), then factoring \(2x^{2}+5x - 12\) as \((2x - 3)(x + 4)\) (wait, let's check: \((2x - 3)(x + 4)=2x^{2}+8x - 3x - 12=2x^{2}+5x - 12\), correct. So denominator is \(x(2x - 3)(x + 4)\). Wait, but let's check the original function: \(f(x)=\frac{2x}{2x^{3}+5x^{2}-12x}\). We can also factor numerator and denominator: numerator is \(2x\), denominator is \(x(2x^{2}+5x - 12)\). So we can cancel the \(x\) (for \(x eq0\)), but we still need to consider the values that make the original denominator zero, including \(x = 0\), and the roots of \(2x^{2}+5x - 12 = 0\). Wait, but let's solve \(2x^{2}+5x - 12 = 0\). Using quadratic formula \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\), where \(a = 2\), \(b = 5\), \(c=-12\). So \(x=\frac{-5\pm\sqrt{25 + 96}}{4}=\frac{-5\pm\sqrt{121}}{4}=\frac{-5\pm11}{4}\). So \(x=\frac{-5 + 11}{4}=\frac{6}{4}=\frac{3}{2}\), and \(x=\frac{-5 - 11}{4}=\frac{-16}{4}=-4\). Wait, but earlier when we factored \(2x^{2}+5x - 12\) as \((2x - 3)(x + 4)\), then setting to zero: \(2x - 3 = 0\) gives \(x=\frac{3}{2}\), \(x + 4 = 0\) gives \(x=-4\). And the original denominator also has a factor of \(x\) (from factoring out \(x\) from \(2x^{3}+5x^{2}-12x\)), so \(x = 0\) is also a root. Wait, but let's check the original denominator: \(2x^{3}+5x^{2}-12x\). Let's plug \(x = 0\): \(0 + 0 - 0 = 0\), so \(x = 0\) is a root. Then \(x=\frac{3}{2}\): \(2(\frac{3}{2})^{3}+5(\frac{3}{2})^{2}-12(\frac{3}{2})=2(\frac{27}{8})+5(\frac{9}{4})-18=\frac{27}{4}+\frac{45}{4}-18=\frac{72}{4}-18 = 18 - 18 = 0\). \(x=-4\): \(2(-4)^{3}+5(-4)^{2}-12(-4)=2(-64)+5(16)+48=-128 + 80 + 48 = 0\). So the denominator is zero when \(x = 0\), \(x=\frac{3}{2}\), or \(x=-4\)? Wait, no, wait: denominator is \(2x^{3}+5x^{2}-12x\), factored as \(x(2x^{2}+5x - 12)=x(2x - 3)(x + 4)\)? Wait, no, \(2x - 3 = 0\) is \(x=\frac{3}{2}\), \(x + 4 = 0\) is \(x=-4\), and \(x = 0\). So the values that make the denominator zero are \(x = 0\), \(x=\frac{3}{2}\), and \(x=-4\)? Wait, but let's check the answer options. The third option is \(\{x\in\mathbb{R}|x eq0,-\frac{3}{2},4\}\)? Wait, no, wait, maybe I made a mistake in factoring. Wait, let's re - factor the denominator:

\(2x^{3}+5x^{2}-12x=x(2x^{2}+5x - 12)\). Let's factor \(2x^{2}+5x - 12\) again. Let's use the AC method. \(a = 2\), \(c=-12\), \(ac=-24\). We need two numbers that multiply to \(-24\) and add to \(5\). The numbers are \(8\) and \(-3\). So, \(2x^{2}+8x-3x - 12 = 2x(x + 4)-3(x + 4)=(2x - 3)(x + 4)\). So \(2x^{2}+5x - 12=(2x - 3)(x + 4)\), so denominator is \(x(2x - 3)(x + 4)\). So setting denominator to zero: \(x = 0\), \(2x - 3 = 0\Rightarrow x=\frac{3}{2}\), \(x + 4 = 0\Rightarrow x=-4\). Wait, but the answer options have \(-\frac{3}{2}\) and \(4\) in some. Wait, maybe I made a sign error. Let's re - examine the original function: \(f(x)=\frac{2x}{2x^{3}+5x^{2}-12x}\). Let's factor denominator as \(2x^{3}+5x^{2}-12x=x(2x^{2}+5x - 12)\). Let's solve \(2x^{2}+5x - 12 = 0\) again. Quadratic formula: \(x=\frac{-5\pm\sqrt{25+96}}{4}=\frac{-5\pm\sqrt{121}}{4}=\frac{-5\pm11}{4}\). So \(x=\frac{-5 + 11}{4}=\frac{6}{4}=\frac{3}{2}\), \(x=\frac{-5 - 11}{4}=\frac{-16}{4}=-4\). So the roots are \(x = 0\), \(x=\frac{3}{2}\), \(x=-4\). But the answer option third is \(\{x\in\mathbb{R}|x eq0,-\frac{3}{2},4\}\)? Wait, no, maybe I factored the denominator wrong. Wait, let's try factoring out a 2 from the denominator: \(2x^{3}+5x^{2}-12x=x(2x^{2}+5x - 12)\). Wait, maybe the quadratic is \(2x^{2}+5x - 12\), but if we factor it as \((2x + 3)(x - 4)\), let's check: \((2x+3)(x - 4)=2x^{2}-8x+3x - 12=2x^{2}-5x - 12\), which is not correct. Wait, maybe the original denominator is \(2x^{3}+5x^{2}-12x\), and when we factor, we can write it as \(x(2x^{2}+5x - 12)\), and then if we make a mistake in the sign when factoring the quadratic. Wait, let's try to factor \(2x^{2}+5x - 12\) as \((2x + a)(x + b)\), then \(2x^{2}+(a + 2b)x+ab\). So \(a + 2b = 5\) and \(ab=-12\). Let's try \(a = 8\), \(b=- \frac{3}{2}\), no, that's not integer. Wait, earlier we had \(a = 8\), \(b=-3\) for the middle term split. Wait, maybe the problem is in the original function's denominator: maybe it's \(2x^{3}+5x^{2}+12x\)? No, the user wrote \(2x^{3}+5x^{2}-12x\). Wait, let's check the answer options. The third option is \(\{x\in\mathbb{R}|x eq0,-\frac{3}{2},4\}\), the fourth is \(\{x\in\mathbb{R}|x eq0,\frac{3}{2},-4\}\). Ah! I see, I think I made a mistake in the sign when factoring. Wait, let's re - do the quadratic: \(2x^{2}+5x - 12\). Wait, if we have \(2x^{2}+5x - 12\), and we want to factor it as \((2x - 3)(x + 4)\), we get \(2x^{2}+8x-3x - 12=2x^{2}+5x - 12\), correct. So the roots are \(x = 0\), \(x=\frac{3}{2}\), \(x=-4\). So the domain is all real numbers except \(x = 0\), \(x=\frac{3}{2}\), and \(x=-4\), which is the fourth option: \(\{x\in\mathbb{R}|x eq0,\frac{3}{2},-4\}\). Wait, let's check the fourth option: \(\{x\in\mathbb{R}|x eq0,\frac{3}{2},-4\}\). Yes, that's the set of all real numbers except \(x = 0\), \(x=\frac{3}{2}\), and \(x=-4\), which are the values that make the denominator zero.

Step2: Determine the domain

The domain of a rational function is all real numbers except the values that make the denominator zero. So we need to find the values of \(x\) for which \(2x^{3}+5x^{2}-12x = 0\). We factored the denominator as \(x(2x - 3)(x + 4)\) (wait, no, \(2x - 3 = 0\) gives \(x=\frac{3}{2}\), \(x + 4 = 0\) gives \(x=-4\), and \(x = 0\) gives \(x = 0\)). So the values that make the denominator zero are \(x = 0\), \(x=\frac{3}{2}\), and \(x=-4\). Therefore, the domain is all real numbers \(x\) such that \(x eq0\), \(x eq\frac{3}{2}\), and \(x eq - 4\).

Answer:

\(\{x\in\mathbb{R}|x
eq0,\frac{3}{2},-4\}\) (the fourth option)