Question

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

Yes, the function is differentiable at \(x=2\).

Step-by-step solution

Find derivative for the function when \(x

To find the derivative of the function \(\frac{1}{2}x + 1\), use the power rule of differentiation which states that the derivative of \(x^n\) is \(n*x^{n-1}\). The derivative of \(\frac{1}{2}x + 1\) is \(\frac{1}{2}\).

Find derivative for the function when \(x≥2\)

To find the derivative of the function \(\sqrt{2x}\), it would be easier to rewrite \(\sqrt{2x}\) as \((2x)^{1/2}\) to apply the power rule. The derivative of \((2x)^{1/2}\) is \(1/(\sqrt{4x})\).

Evaluate the left limit at \(x=2\)

The left limit of the derivative at \(x=2\) is the value of the derivative as \(x\) approaches \(2\) from the left. Substitute \(x=2\) into the first derivative, \(\frac{1}{2}\), it remains \(\frac{1}{2}\).

Evaluate the right limit at \(x=2\)

The right limit of the derivative at \(x=2\) is the value of the derivative as \(x\) approaches \(2\) from the right. Substitute \(x=2\) into the second derivative, that is \(1/(\sqrt{4x})\), the result is \(\frac{1}{2}\).

Compare the Right and Left Limit

The function is differentiable at \(x=2\) if the right limit and left limit are equal. In this case, as both limits equal to \(\frac{1}{2}\), thus the function is differentiable at \(x=2\).