Question

Maximum value of the function f(x) = 1 - {x} / 3 + {x} is where {.} denotes the fractional part of (x)

Answer 1

Answer:

The maximum value of fractional part of x tends to 1 but is never 1 actually. Its lowest value is 0 though.

f

′

(x)=

(1+{x})

2

1

.

This is always positive. So f(x) is an increasing function.

Hence minimum value of the function is at x=0. f(x)=0

Maximum value of f(x) is as {x} tends to 1. The limiting value of the fuction is

2

1

but it is never actually attained by the function

Answer 2

Answer:

The maximum value of the function is 1

Step-by-step explanation:

The given function is

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

We can rearrange it in the following way

$f(x) = \frac{3 + \left\{ x\right\} - \left\{ x\right\}}{3+\left\{ x\right\} } = \frac{3}{3 + \left\{ x\right\} }$

We have to find the maximum value of the above function.

For any function or expression to be maximum,

its denominator must be minimum.

Hence,

for our function 3+{x} must be minimum.

The range of {x} is

$0 \leqslant \left\{ x\right\} < 1$

Hence its minimum value is zero.

Therefore,

the minimum value of 3+{x} is

$3 + 0 = 3$

Hence,the maximum value of our expression becomes

$\frac{3}{3 + 0} = 1$

Therefore,

the maximum value of the function is 1

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