Question

Determining Differentiability In Exercises 89 and \(90,\) determine whether the function is differentiable at \(x=2\) . $$ f(x)=\left\\{\begin{array}{ll}{\frac{1}{2} x+1,} & {x<2} \\ {\sqrt{2 x},} & {x \geq 2}\end{array}\right. $$

Short answer

No, the given function is not differentiable at \(x=2\).

Step-by-step solution

Compute the left-hand derivative at x=2

The left-hand derivative, denoted as \(f'(2^-)\), is the derivative of the function for values of x less than 2. Hence we should consider the first part of the piecewise function \(\frac{1}{2}x + 1\) to compute the derivative, which is \(\frac{1}{2}\). Now substitute \(x=2\) to calculate \(f'(2^-)\) which equals to \(\frac{1}{2}.\)

Compute the right-hand derivative at x=2

The right-hand derivative, denoted as \(f'(2^+)\), is the derivative of the function for values of x greater than or equal to 2. Hence we should consider the second part of the piecewise function \(\sqrt{2x}\) to compute the derivative. The derivative of \(\sqrt{2x}\) is \(\frac{1}{\sqrt{x}}\). After substituting \(x=2\), \(f'(2^+)=\frac{1}{\sqrt{2}}.\)

Comparing the left and right hand derivatives

If the left-hand derivative and the right-hand derivative at x=2 are equal, then the function is differentiable at x=2. In this case, \(f'(2^-) = \frac{1}{2}\) and \(f'(2^+) =\frac{1}{\sqrt{2}}\). Since these two values are not equal, the function is not differentiable at x=2.

Key concepts

Piecewise Functions

When studying functions, we often encounter piecewise functions, which are defined differently for different intervals of the input variable, usually depicted as 'x'. These functions can appear complex at first, but they are simply a combination of two or more functions, each applied to a certain interval or 'piece' of the domain. To write down a piecewise function, you list each function and the corresponding interval next to it, typically using curly brackets to group them together.

For example, if we have a function f(x) that equals x + 2 when x is less than 3, and equals 2x when x is greater than or equal to 3, then this can be expressed as follows:
\[\begin{equation}f(x) = \left\{\begin{array}{ll}{x + 2,} & {x<3} \ {2x,} &{x \geq 3}\end{array}\right.\end{equation}\]
Determining the properties, such as continuity and differentiability, of piecewise functions often requires special attention at the points where the function 'pieces' join, known as transition points or breakpoints. In our exercise, we are particularly interested in the behavior of a piecewise function at x = 2.

Left-Hand Derivative

The left-hand derivative of a function at a certain point x = a concerns the slope of the function as we approach a from values less than a. It gives us information about how the function changes momentarily as we move towards a from the left side.

To calculate the left-hand derivative, also denoted as f'(a^-), we take the derivative of the part of the piecewise function applicable to the interval just left of x = a, and then evaluate this derivative at a. In the case of our exercise, the left-hand derivative at x = 2 is obtained from the part of the function f(x) where x is less than 2, which is the linear part. The result is a constant slope, representing the fact that the function is changing at a steady rate as it approaches the point from the left. Finding the left-hand derivative is crucial when assessing the differentiability of a piecewise function at its transition point.

Right-Hand Derivative

The concept of the right-hand derivative parallels that of the left-hand derivative but focuses on the behavior of the function as it approaches a point x = a from the right, meaning from values greater than a. This is denoted as f'(a^+) and examines the instantaneous rate of change of the function as we move towards the point from the right.

To calculate this, we look at the derivative of the part of the piecewise function that is relevant for the interval just right of x = a, and then evaluate this derivative right at a. In our exercise, when x is greater than or equal to 2, we have a square root function. Upon differentiating and substituting x = 2 into this derivative, we find the slope of the function as it approaches the point from the right. The comparison of the right-hand and left-hand derivatives at the transition point tells us whether the function can smoothly transition from one part to the other, which is a critical aspect of assessing differentiability.