Answer to Question #308450 in Real Analysis for Pankaj

Question

Determine the points of discontinuity of the function f and the nature of discontinuity at each of those points:

f ={-x², when x ≤ 0.

4-5x , when 0<x≤1

3x-4x², when 1<x≤2

-12x + 2x , when x>2}

Also check whether the function f is derivable at x = 1

Answer 1

$f(x)= \begin{cases} -x² &\text{if } x≤0 \\ 4-5x&\text{if } 0<x≤1\\ 3x-4x²&\text{if } 1<x≤2\\ -12x+2x &\text{if } x≥2 \end{cases}$

f(x) is polynomial function on every intervals (-∞,0), (0,1), (1,2) and (2,∞).

As polynomial function is always continuous function, here f(x) can have only discontinuites at x=0, 1, 2.

Let us check the continuity at these points.

At x=0

f(0-0) = $\lim_{x\to 0^-}{(-x²)}=0$

f(0+0) =$\lim_{x\to 0^+}{(4-5x)}=4$

f(0) = -0² = 0

Since f(0-0) ≠ f(0+0) , f(x) is discontinuous at x= 0

At x = 1

f(1-0) = $\lim_{x\to 1^-}{(4-5x)}=-1$

f(1+0) = $\lim_{x\to1^+}(3x-4x²)=-1$

f(1) = 4-5 = -1

Since f(1-0) =f(1)= f(1+0) , f(x) is continuous at x= 1

At x = 2

f(2-0) = $\lim_{x\to 2^-}{(3x-4x²)}=-10$

f(2+0) = $\lim_{x\to2^+}(-12x+2x)=-20$

f(2) = -24+4= -20

Since f(2-0)≠f(2+0), f(x) is discontinuous at x= 2

So points of discontinuity are x=0 and x=2