Question

Tangent lines Determine whether the graph of \(y=x^{\sqrt{x}}\) has any horizontal tangent lines.

Short answer

Answer: No, there are no horizontal tangent lines for the given function.

Step-by-step solution

Rewrite the function

First, we rewrite the function in the form \(y=x^{\frac{1}{2}x}\) so that we can apply the chain rule later.

Apply Chain Rule

To find the derivative \(y'\), we can apply the chain rule: \((u^v)' = vu^{v-1}u'\) Here, \(u=x\) and \(v=\frac{1}{2}x\) First, find \(u' = \frac{du}{dx}\). Since \(u=x\), \(u' = 1\) Then find \(v' = \frac{dv}{dx}\). Since \(v=\frac{1}{2}x\), \(v' = \frac{1}{2}\) Now we can calculate \(y'\) by plugging \(u\), \(v\), \(u'\), and \(v'\) into the chain rule formula.

Calculate the derivative

Using the chain rule formula, let's plug in the values we found before: \(y' = \big(vu^{v-1}\big)u' + \big(v' u^v\big)\) Plug in the values, \(u = x\), \(v = \frac{1}{2}x\), \(u' = 1\), and \(v' = \frac{1}{2}\): \(y' = \left(\frac{1}{2}x \cdot x^{\frac{1}{2}x - 1}\right) \cdot 1 + \left(\frac{1}{2} \cdot x^{\frac{1}{2}x}\right)\) Now, simplify the expression: \(y' = \frac{1}{2}x^{\frac{1}{2}x - 1} + \frac{1}{2}x^{\frac{1}{2}x}\)

Check for horizontal tangent lines

A horizontal tangent line occurs when \(y' = 0\). To check, let's set the derivative equal to zero and try to solve for x: \(0 = \frac{1}{2}x^{\frac{1}{2}x - 1} + \frac{1}{2}x^{\frac{1}{2}x}\) It's hard to find an analytic solution for it, but we can examine the behavior of the function. As \(x\) approaches \(0\), the first term \(\frac{1}{2}x^{\frac{1}{2}x - 1}\) approaches infinity, while the second term \(\frac{1}{2}x^{\frac{1}{2}x}\) approaches \(1\). Thus, there is no horizontal tangent in this region. As \(x\) goes to infinity, both terms approach infinity, so there is no horizontal tangent in that region either. Therefore, there are no horizontal tangent lines for the function \(y=x^{\sqrt{x}}\).

Key concepts

Chain Rule

The chain rule is a fundamental tool in calculus used to find the derivative of composite functions. This rule is essential when dealing with functions composed of multiple sub-functions layered on top of each other. In simpler terms, the chain rule allows us to differentiate a function that is inside another function. For instance, if we have a composite function like \(y = (x^2 + 3)^{4}\), the chain rule helps us differentiate it by treating the inner function \((x^2 + 3)\) and the outer function \((z^4)\).
The general formula for the chain rule is \( (f(g(x)))' = f'(g(x)) \cdot g'(x) \). In our given exercise involving \(y = x^{\sqrt{x}}\), the chain rule enables us to find the derivative by detailing each part of the composite function: \(u = x\) and \(v = \frac{1}{2}x\).
Using this method, the chain rule results in: \((u^v)' = vu^{v-1}u' + v'u^v \), where \(u' = 1\) and \(v' = \frac{1}{2}\). This application of the chain rule is essential for subsequent steps in the solution.

Derivative Calculation

Calculating the derivative accurately is crucial to understanding how a function behaves, especially at specific points. When the exercise initially presents us with the function \(y = x^{\sqrt{x}}\), simplifying it to \(y = x^{\frac{1}{2}x}\) helps us better manage the operation needed.
We first find the derivative using the chain rule we discussed earlier. Here, by substituting the relevant values, we derive the formula: \(y' = \left(\frac{1}{2}x \cdot x^{\frac{1}{2}x - 1}\right) \cdot 1 + \left(\frac{1}{2} \cdot x^{\frac{1}{2}x}\right)\).
This formula reveals how y changes as x changes: the derivative. Simplifying this expression gives us the final derivative \(y' = \frac{1}{2}x^{\frac{1}{2}x - 1} + \frac{1}{2}x^{\frac{1}{2}x}\), which serves as the foundation for analyzing the function's behavior further.

Horizontal Tangents

A horizontal tangent occurs when the slope of the tangent line to a graph is zero, meaning that the change in y with respect to x is zero at that point. This scenario arises when the derivative of the function equals zero.
For the function \(y = x^{\sqrt{x}}\), finding horizontal tangents involves setting the derivative \(y' = \frac{1}{2}x^{\frac{1}{2}x - 1} + \frac{1}{2}x^{\frac{1}{2}x}\) to zero. Solving this equation analytically can be difficult due to its complexity, making graphical or numerical methods useful alternatives. However, theoretical evaluations indicate:

• As \(x\) approaches zero, one term trends towards infinity while the other approaches one.
• As \(x\) goes to infinity, both terms tend toward infinity.

With these insights, no horizontal tangent lines exist for \(y = x^{\sqrt{x}}\) across its domain. This conclusion stems from the unbounded growth of both components of \(y'\) as \(x\) varies, showing no single point where \(y'\) is zero.