Inspect the graph of the function to determine whet A square root function, \( n(x)=\sqrt{3 x} \), on \( (0, \infty) \) On the interval \( (0, \infty), n(x)=\sqrt{3 x} \) is concave neither co

1. We are given the function \[ n(x)=\sqrt{3x} \] on the interval \((0, \infty)\). 2. To determine concavity, we compute the second derivative \(n''(x)\). 3. First, differentiate \(n(x)\): \[ n(x)=\sqrt{3x} = (3x)^{\frac{1}{2}}. \] Using the chain rule, \[ n'(x)=\frac{1}{2}(3x)^{-\frac{1}{2}}(3)=\frac{3}{2}(3x)^{-\frac{1}{2}}. \] Rewrite \((3x)^{-\frac{1}{2}}\) as: \[ n'(x)=\frac{3}{2\sqrt{3x}}=\frac{\sqrt{3}}{2\sqrt{x}}. \] 4. Next, differentiate \(n'(x)\) to find the second derivative: \[ n'(x)=\frac{\sqrt{3}}{2} x^{-\frac{1}{2}}. \] Differentiating with respect to \(x\), \[ n''(x)=\frac{\sqrt{3}}{2}\left(-\frac{1}{2}\right)x^{-\frac{3}{2}} = -\frac{\sqrt{3}}{4}x^{-\frac{3}{2}}. \] 5. Since \(x^{-\frac{3}{2}}>0\) for \(x>0\), we have \[ n''(x)=-\frac{\sqrt{3}}{4}x^{-\frac{3}{2}} < 0 \] for all \(x>0\). 6. Because the second derivative is negative on \((0, \infty)\), the function \(n(x)=\sqrt{3x}\) is concave (concave downward) on this interval.