Let f:RR be a function defined as f(x)=cases5, & x1+bx, & 1

Question

Let $f:\mathbb{R}	o\mathbb{R}$ be a function defined as $f(x)=\begin{cases}5, & x\le1\a+bx, & 1<x<3\b+5x, & 3\le x<5\30, & x\ge5\end{cases}$. Then, $f$ is:

Accepted Answer

Checking

if f(x) is continuous at x = 1 :

f(1-) = 5

f(1) = 5

f(1+) = a + b

if f(x) is continuous at x = 1,

then

f(1-) = f(1) = f(1+)

$$ \Rightarrow $$  5 = 5 = a + b

$$ \therefore $$  a + b = 5 . . . . . . . .  (1)

checking if f(x) is continuous at x = 3 :

f(3-) = a + 3b

f(3) = b + 15

f(3+) = b + 15

if   f(x) = is continuous at x = 3

then,

f(3-) = f(3) = f(3+)

$$ \Rightarrow $$  a + 3b = b + 15 = b + 15

$$ \Rightarrow $$  a + 2b = 15 . . . . . (2)

checking if f(x) is continuous at x = 5 :

f(5-) = b + 25

f(5) = 30

f(5+) = 30

if f(x) is continuous at x = 5 then,

f(5-) = f(5) = f(5+)

$$ \Rightarrow $$   b + 25 = 30 = 30

$$ \Rightarrow $$   b = 5

By putting this value in equation (2),  we get,

a + 2(5) = 15

$$ \Rightarrow $$  a = 5

when a = 5 and b = 5 then equation (1)

a + b = 5 does not satisfy.

$$ \therefore $$  f is not continuous for any value of a and b.

Solutions · 1