Find the intervals in which the function f (x) = log left parenthesis 1 plus straight x right parenthesis minus fraction numerator 2 straight x over denominator 1 plus straight x end fraction is increasing or decreasing

Asked by Topperlearning User | 6th Aug, 2014, 10:12: AM

Expert Answer:

We have,

f (x) = log left parenthesis 1 plus straight x right parenthesis minus fraction numerator 2 straight x over denominator 1 plus straight x end fraction

Clearly, f (x) is defined for all x satisfying

x + 1 > 0 i.e. x > – 1

So, domain (f) = (–1, ¥)

Now,

f (x) = log left parenthesis 1 plus straight x right parenthesis minus fraction numerator 2 straight x over denominator 1 plus straight x end fraction

Þ f ¢ (x) = fraction numerator 1 over denominator 1 plus straight x end fraction. straight d over dx left parenthesis straight x plus 1 right parenthesis minus fraction numerator left parenthesis 1 plus straight x right parenthesis.2 minus 2 straight x left parenthesis 0 plus 1 right parenthesis over denominator left parenthesis 1 plus straight x right parenthesis squared end fraction

Þ f ¢ (x) = fraction numerator 1 over denominator 1 plus straight x end fraction minus fraction numerator 2 over denominator left parenthesis 1 plus straight x right parenthesis squared end fraction

Þ f ¢ (x) = fraction numerator left parenthesis 1 plus straight x right parenthesis minus 2 over denominator left parenthesis 1 plus straight x right parenthesis squared end fraction equals fraction numerator straight x minus 1 over denominator left parenthesis straight x plus 1 right parenthesis squared end fraction

For f (x) is be increasing, we must have

f¢ (x) > 0

Þ fraction numerator left parenthesis straight x minus 1 right parenthesis over denominator left parenthesis straight x plus 1 right parenthesis squared end fraction greater than 0 space space space open square brackets because open parentheses fraction numerator 1 over denominator 1 plus straight x end fraction close parentheses squared greater than 0 close square brackets

Þ x - 1 > 0

Þ x > 1

Þ x Î (1, ¥)

So, f (x) is increasing on (1, ¥)

Hence, f (x) is increasing in its domain

Answered by  | 6th Aug, 2014, 12:12: PM