discuss the curve of y=ln(1+x^2)?

Answer 1

Well...

It's symmetric about the y-axis, because x² = (-x)².

At x = 0, y = ln 1 = 0

As x gets bigger and bigger, it looks more like the graph of

y = ln x² = 2ln x

So it increases in both directions but very very slowly.

Is that what you're looking for?

Answer 2

To discuss the curve, we likely want: Domain, Range, x and y intercepts, asymptotes, intervals of increase/decrease, local minima/maxima, points of inflection, intervals of inflection. If that's the case, then so be it.

First, let's solve for the intercepts. To find the y-intercept, make x equal to 0.

y = ln (1 + 0^2) = ln1 = 0, so 0 is a y intercept

To find the x-intercept, make y equal to 0
0 = ln(1 + x^2)
e^0 = 1+x^2
1 = 1+x^2, x = 0, so 0 is an x-intercept

There aren't any horizontal or vertical asymptotes.

y' = 1/(1+x^2) [2x]
y' = 2x/(1 + x^2)

Critical numbers: x = 0
increasing on (-infinity, 0) U (0, infinity)
No local extrema.

y'' = [2(1+x^2) - 2x(2x)]/[1+x^2]^2
y'' = [2 + 2x^2 - 4x^2] / [1+x^2]^2
y'' = [2 - 2x^2] / [1+x^2]^2

Therefore, if we make that 0, then 2 - 2x^2 = 0, and
x^2 = 1, so x = -1, 1
concave down on (-infinity, -1) U (1, infinity)
concave up on (-1, 1)

Points of inflection: f(-1) = ln(2), f(1) = ln(2)
Inflection points at (-1, ln2) and (1, ln(2))

Domain: All real numbers.
Range: All real numbers.

Answer 3

Discuss is a very vague term. I agree with most of what was written above, except that 0 is a local minima, and the range is the nonnegative real numbers only, since 1+x^2 >= 1, ln(1)=0 and ln is monotonically increasing.