Question

f(x)=\frac{1}{2}\sqrt3{x - 5}-2

the function has an inflection point at (5, -2).

the function has a range of (-\infty, \infty).

as x \to \infty, f(x) \to -\infty

Explanation:

Step1: Analyze the inflection point

For a function \(y = f(x)\), an inflection point is a point where the concavity changes. The general form of a cube - root function is \(y = a\sqrt[3]{x - h}+k\). The second - derivative of \(y=\frac{1}{2}\sqrt[3]{x - 5}-2\) is non - zero and the function \(y = \sqrt[3]{x}\) has no inflection point in the traditional sense (since its second - derivative \(y''=\frac{2}{9}x^{-\frac{5}{3}}\) is never zero for \(x
eq0\) and has a vertical tangent at \(x = 0\)). For \(y=\frac{1}{2}\sqrt[3]{x - 5}-2\), we can rewrite it as \(y=\frac{1}{2}(x - 5)^{\frac{1}{3}}-2\). The first - derivative \(y'=\frac{1}{6}(x - 5)^{-\frac{2}{3}}\) and the second - derivative \(y''=-\frac{1}{9}(x - 5)^{-\frac{5}{3}}\). The second - derivative is undefined at \(x = 5\), but there is no change in concavity across \(x = 5\) (the concavity is negative for \(x>5\) and positive for \(x<5\) in terms of the sign of \(y''\), but since the function is not smooth at the vertical tangent - like behavior at \(x = 5\), we don't call it an inflection point in the strict calculus sense (where the function is twice - differentiable in an open interval around the point)).

Step2: Analyze the range

The range of the basic cube - root function \(y=\sqrt[3]{u}\) is \((-\infty,\infty)\). For the function \(y=\frac{1}{2}\sqrt[3]{x - 5}-2\), let \(u=x - 5\). As \(u\) takes all real values (since \(x\in(-\infty,\infty)\)), \(\sqrt[3]{u}\) takes all real values. Then \(y=\frac{1}{2}\sqrt[3]{u}-2\). If \(z=\sqrt[3]{u}\), then \(y=\frac{1}{2}z-2\). Since \(z\in(-\infty,\infty)\), when \(z\to-\infty\), \(y=\frac{1}{2}z - 2\to-\infty\) and when \(z\to\infty\), \(y=\frac{1}{2}z-2\to\infty\). So the range of \(y = f(x)\) is \((-\infty,\infty)\).

Step3: Analyze the end - behavior

As \(x\to\infty\), let \(u=x - 5\to\infty\). Then \(y=\frac{1}{2}\sqrt[3]{x - 5}-2\). Since \(\sqrt[3]{x-5}\to\infty\) as \(x\to\infty\), then \(y=\frac{1}{2}\sqrt[3]{x - 5}-2\to\infty\) (because \(y=\frac{1}{2}z-2\) with \(z=\sqrt[3]{x - 5}\) and \(z\to\infty\) as \(x\to\infty\)).

Answer:

1. False
2. True
3. False