Question

How do you use the first and second derivatives to sketch `f(x)=x^2/(x^2+5)`?

Answer 1

See the explanation below

graph{x^2/(x^2+5) [-5, 5, -2.5, 2.5]}

Explanation:

First we note that, as `x^2 + 5 > 0` for all `x`, the function is defined and continuous in all of `RR`.

We also note that as the numerator and denominator only contain terms of even degree in `x` the function is even, that is:

`f(-x) = f(x)`

so the graph will be symmetric with respect to the `y` axis.

We then evaluate the behavior at the limits of the domain:

`lim_(x->-oo) x^2/(x^2+5) = lim_(x->+oo) x^2/(x^2+5) = 1`

so on both sides the line `y = 1` is an asymptote.

Now we calculate the first derivative using the quotient rule:

`f'(x) = (2x(x^2+5) - 2x^3)/(x^2+5)^2 = (10x)/(x^2+5)^2`

The only critical point is then `x=0` and we have:

`{(f'(x) < 0 " for " x < 0),(f'(x) > 0 " for " x > 0):}`

so `f(x)` is decreasing in `(-oo,0)` and increasing in `(0,+oo)`, which means the value for `x=0` is a local minimum.

As the value of the minimum is `f(0) = 0` and the function is positive everywhere else, this is also an absolute minimum.

Calculating the second derivative:

`f''(x) = (10(x^2+5)^2 - 40x^2(x^2+5))/(x^2+5)^4 = (10x^2+50-40x^2)/(x^2+5)^3 =-10(3x^2-5)/(x^2+5)^3`

The points of inflection are then the roots of:

`3x^2-5 = 0`

`x=+-sqrt(5/3)`

to solve the inequation `f''(x) > 0` we can then note that:

(i) the denominator is always positive

(ii) the numerator `3x^2-5` is a second order polynomial with two roots, so it is negative in the interval between the roots and positive outside.

(iii) the sign of `f''(x)` is the opposite of the numerator's.

Then:

`f(x)` is concave down for `x in (-oo, -sqrt(5/3))`

`f(x)` is concave up for `x in (-sqrt(5/3),sqrt(5/3))`

`f(x)` is concave down for `x in (sqrt(5/3),+oo)`

Answer 2

A few additional graphing notes

Explanation:

In addition to the excellent answer that you see earlier in this thread, I would like to add a few (general) graphing tips.

Graphing techniques in Algebra provide an excellent idea of what the graph looks like. Calculus only provides specific details about relative extremes and concavity.

The graph of a rational function may change from increasing to decreasing (or vice-versa) only at a critical value -- a relative extreme, removable discontinuity, or vertical asymptote. You can already determine much of this information with algebra, but you will not know the locations of relative extremes (maxes or mins) without Calculus.

The graph of a rational function may change concavity only at an inflection point. You will not know these values without Calculus.