Question

Determine the point(s) at which the graph of \(f(x)=\frac{x}{\sqrt{2 x-1}}\) has a horizontal tangent line.

Short answer

The function \(f(x) = \frac{x}{\sqrt{2x-1}}\) has no point at which it has a horizontal tangent line.

Step-by-step solution

Find the derivative of \(f(x)\)

Using the quotient rule, the derivative of the function \(f(x) = \frac{x}{\sqrt{2x-1}}\) becomes \(f'(x) = \frac{\sqrt{2x-1} - (x \cdot \frac{1}{2 \sqrt{2x-1}})}{(2x-1)}\). Simplifying, we get \(f'(x) = \frac{1}{2 \sqrt{2x-1}}\).

Set the derivative equal to 0 and solve for \(x\)

Setting \(f'(x) = 0\), we get \(\frac{1}{2 \sqrt{2x-1}} = 0\). This equation doesn't have a real root, hence there are no values of \(x\) for which \(f(x)\) has a horizontal tangent.

Check for points of undefined derivative as potential horizontal tangents

The derivative is undefined when the denominator equals zero. But solving \(2x-1 = 0\) for \(x\) gives \(x = 0.5\), which is not in the domain of the original function \(f(x)\) because for \(x = 0.5\), \(f(x)\) is undefined. Therefore, it is not considered as a point of horizontal tangent line.