Question

Find the one-sided derivatives of the function f(x) = |x+ 18|at the point x= -18 if they exist. If the derivative does not exist write DNE for your answer.

Answer 1

Answer:

Left-hand derivative at x = -18 is -1

Right-hand derivative at x = -18 is 1

Explanation:

Given the function;

$f(x)=|x+18|$

We'll use the below formula to find the left-hand derivative of the above function;

$f^{\prime}(a^-)=\lim _{h\to0^-}\frac{f(a+h)-f(a)}{h}$

If we substitute and solve, we'll have;

$\begin{gathered} f^{\prime}(a^-)=\lim _{h\to0^-}=\frac{|(-18+h)+18|-f(-18)}{h} \\ =\lim _{h\to0^-}=\frac{|(-18+h)+18|-0}{h} \\ =\lim _{h\to0^-}=\frac{|h|}{h} \end{gathered}$

Since this is a left-hand derivative, therefore h < 0;

$\lim _{h\to0^-}=\frac{-h}{h}=-1$

Let's go ahead and determine the right-hand derivative using the below formula;

$\begin{gathered} f^{\prime}(a^+)=\lim _{h\to0^+}\frac{f(a+h)-f(a)}{h} \\ f^{\prime}(a^+)=\lim _{h\to0^+}=\frac{|(-18+h)+18|-f(-18)}{h} \\ =\lim _{h\to0^+}=\frac{|(-18+h)+18|-0}{h} \\ =\lim _{h\to0^+}=\frac{|h|}{h} \end{gathered}$

Since this is a right-hand derivative, so h > 0;

$\lim _{h\to0^+}=\frac{h}{h}=1$

For a function to be differentiable at any point, its left-hand and right-hand derivative must exist and they must coincide.

From the above, we have that the left-hand derivative = -1 and the right-hand derivative = 1.