Question

check the differentiability of f(x) in [-2,2]
if f(x) is (see attachment). ​

Answer 1

The function f(x) in [-2, 2] is defined as,

$\small\text{ \longrightarrow f(x)=\left\{\begin{array}{lr}\left(\cos\left(\dfrac{\pi}{2}\right)\right)\big(|x|-\{x\}\big),&x<1\\\\\{x\}\,\sqrt{4x^2-12x+9},&x\geq1\end{array}\right. }$

Including the interval [-2, 2] in the definition,

$\small\text{ \longrightarrow f(x)=\left\{\begin{array}{lr}\left(\cos\left(\dfrac{\pi}{2}\right)\right)\big(|x|-\{x\}\big),&-2\leq x<1\\\\\{x\}\,\sqrt{4x^2-12x+9},&1\leq x\leq2\end{array}\right. }$

Since $\small\cos\left(\dfrac{\pi}{2}\right)=0,$

$\small\text{ \longrightarrow f(x)=\left\{\begin{array}{lr}0,&-2\leq x<1\\\\\{x\}\sqrt{4x^2-12x+9},&1\leq x\leq2\end{array}\right. }$

We see that,

$\small\longrightarrow\sqrt{4x^2-12x+9}=\sqrt{(2x)^2-2\cdot2x\cdot3+3^2}$

$\small\longrightarrow\sqrt{4x^2-12x+9}=\sqrt{(2x-3)^2}$

$\small\longrightarrow\sqrt{4x^2-12x+9}=|2x-3|$

$\small\text{ \longrightarrow\sqrt{4x^2-12x+9}=\left\{\begin{array}{lr}3-2x,&x\leq\dfrac{3}{2}\\\\2x-3,&x\geq\dfrac{3}{2}\end{array}\right. }$

Then, taking {x} = x - [x] also, we get,

$\small\text{ \longrightarrow f(x)=\left\{\begin{array}{lr}0,&-2\leq x<1\\\\\big(x-[x]\big)(3-2x),&1\leq x\leq\dfrac{3}{2}\\\\\big(x-[x]\big)(2x-3),&\dfrac{3}{2}\leq x\leq2\end{array}\right. }$

Since [x] = 1 for 1 ≤ x < 2 and [x] = 2 for x = 2, the definition becomes,

$\small\text{ \longrightarrow f(x)=\left\{\begin{array}{lr}0,&-2\leq x<1\\\\(x-1)(3-2x),&1\leq x\leq\dfrac{3}{2}\\\\(x-1)(2x-3),&\dfrac{3}{2}\leq x<2\\\\(x-2)(2x-3),&x=2\end{array}\right. }$

We see f(x) = 0 at x = 2.

$\small\text{ \longrightarrow f(x)=\left\{\begin{array}{lr}0,&-2\leq x<1\\\\-2x^2+5x-3,&1\leq x\leq\dfrac{3}{2}\\\\2x^2-5x+3,&\dfrac{3}{2}\leq x<2\\\\0,&x=2\end{array}\right. }$

Now the derivative is,

$\small\text{ \longrightarrow f'(x)=\left\{\begin{array}{lr}0,&-2\leq x<1\\\\-4x+5,&1\leq x\leq\dfrac{3}{2}\\\\4x-5,&\dfrac{3}{2}\leq x<2\\\\0,&x=2\end{array}\right. }$

We see the following.

• $\small\displaystyle\lim_{x\to1^-}f'(x)=0\neq1=\lim_{x\to1^+}f'(x)$

• $\small\text{ \displaystyle\lim_{x\to\frac{3}{2}^-}f'(x)=-1\neq1=\lim_{x\to\frac{3}{2}^+}f'(x) }$

• $\small\displaystyle\lim_{x\to2^-}f'(x)=3\neq0=\lim_{x\to2^+}f'(x)$

It means f(x) is not differentiable at points x = 1, x = 3/2, x = 2.