Question

Which of the following statement is true about the function

f(x) = 3[x] + 2

Answer 1

Answer:

f

′

(x) changes its sign twice as x varies from (−∞,∞)

For x<0

f(x)=

3

x

3

−4x

Hence

f

′

(x)=x

2

−4

Now for monotonically increasing function

f

′

(x)>0

Or

xϵ(−∞,−2)∪(2,∞). However domain is x<0.

Hence xϵ(−∞.−2).

For

0<x<1

f(x)=x

3

.

f

′

(x)=3x

2

Now for monotonically increasing function f

′

(x)>0

Or

x>0.

Hence f(x) is an increasing function in (0,1).

For x>1

f(x)=

x

f

′

(x)=

2

2

1

for monotonically increasing function f

′

(x)>0

Hence

x>0.

Thus summing up, we get f(x) as an increasing function in the interval of

(−∞,−2)∪(0,∞).

Hence it changes sign twice, once at x=2 and another at x=0.