Question

F:R—>R, f(x)=log(1+x²),Determine intervals in which the given function are strictly increasing or strictly decreasing.

Answer 1

f : R --------> R , f(x) = log(1 + x²)

f(x) = log(1 + x²)

differentiate with respect to x,

f'(x) = 1/(1 + x²) × 2x

f'(x) = 2x/(1 + x²)

now, f'(x) = 0

2x/(1 + x²) = 0 => x = 0


case 1 :- x > 0, f'(x) > 0
so, function is strictly increasing in (0, ∞)

case 2 :- x < 0, f'(x) < 0
so, function is strictly decreasing in (-∞, 0)

therefore, function is strictly increasing in (0, ∞) while function is strictly decreasing in (-∞, 0).

Answer 2

Dear student:

Given: F:R—>R,

f(x)=log(1+x²)

For determining the intervals in which f is increasing and decreasing.

Find derivative of f(x)

Then see the derivative in which f is positive and negative.

If it is positive then f is increasing

And if it is negative then f is decreasing.

See the attachment.