Question

The function f: R→R defined by f(x)= | x-1|+ | 3-x| is differentiable at x= 4.

True or false with full explanation

Answer 1

The given function is $f(x)=|x-1|+|3-x|,\ x\in\R.$

But, the derivative of a sum is the sum of derivatives so;

$\displaystyle \frac{d}{dx} \left(\left|{x - 3}\right| + \left|{x - 1}\right|\right) = \frac{d}{dx} \left(\left|{x - 3}\right|\right) + \frac{d}{dx} \left(\left|{x - 1}\right|\right)$

The function $\displaystyle \left|{x - 3}\right|$ is the composition $\displaystyle f{\left(g{\left(x \right)} \right)}$ of two functions $\displaystyle f{\left(u \right)} = \left|{u}\right|$ and

$\displaystyle g{\left(x \right)} = x - 3$.

Also, The function $\displaystyle \left|{x - 1}\right|$ is the composition  $\displaystyle h{\left(k{\left(x \right)} \right)}$ of two functions $\displaystyle h{\left(v \right)} = \left|{v}\right|$ and

$\displaystyle k{\left(x \right)} = x - 1$

Now, apply the chain rule which is $\displaystyle \frac{d}{dx} \left(f{\left(g{\left(x \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dx} \left(g{\left(x \right)}\right)$ and

$\displaystyle \frac{d}{dx} \left(h{\left(k{\left(x \right)} \right)}\right) = \frac{d}{du} \left(h{\left(v \right)}\right) \frac{d}{dx} \left(k{\left(x \right)}\right)$.

Where $\displaystyle \frac{d}{du} \left(\left|{u}\right|\right) = \frac{u}{\left|{u}\right|}\text{ and } \frac{d}{dv} \left(\left|{v}\right|\right) = \frac{v}{\left|{v}\right|}$

We have;

$\displaystyle \frac{d}{dx} \left(\left|{x - 3}\right|\right) + \frac{d}{dx} \left(\left|{x - 1}\right|\right) = \frac{d}{du} \left(\left|{u}\right|\right) \frac{d}{dx} \left(x - 3\right) + \frac{d}{dv}(|v|)\frac{d}{dx} \left({x - 1}\right)$

$\displaystyle =\frac{\left(u\right) \frac{d}{dx} \left(x - 3\right)}{\left|u\right|} + \frac{\left(v\right) \frac{d}{dx} \left(x - 1\right)}{\left|v\right|} \\$

$\displaystyle =\frac{\left(x - 3\right) \frac{d}{dx} \left(x - 3\right)}{\left|{x - 3}\right|} + \frac{\left(x - 1\right) \frac{d}{dx} \left(x - 1\right)}{\left|{x - 1}\right|} \\$

$\displaystyle = \frac{\left(x - 3\right) }{\left|{x - 3}\right|} + \frac{x - 1}{\left|{x - 1}\right|}$

Thus,

$\displaystyle \frac{d}{dx} \left(\left|{x - 3}\right| + \left|{x - 1}\right|\right) = \frac{x - 3}{\left|{x - 3}\right|} + \frac{x - 1}{\left|{x - 1}\right|}$

Hence, the given function is differentiable at $\displaystyle x=4$ and

$\displaystyle \frac{d}{dx} \left(\left|{x - 3}\right| + \left|{x - 1}\right|\right)|_{\left(x = 4\right)} = 2$