Question

How do you use the first and second derivatives to sketch `f(x) = sqrt(4 - x^2)`?

Answer 1

The domain of the function is the interval `x in [-2,2]`.

We can note that the function is even, that is `f(-x) = f(x)` so its graph will be symmetrical with respect to the `y` axis.

We can also note that `f(x) >=0` and `f(-2) = f(2) = 0`.

Calculate now:

`(df)/dx = -x/sqrt(4-x^2)`

which means that the function is differentiable only in the interior of the interval.

As:

`lim_(x->-2^+) (df)/dx = +oo`

`lim_(x->2^-) (df)/dx = -oo`

the tangent to the graph at the limits of the interval of definition is vertical.

The only critical point where:

`(df)/dx = 0`

is `x=0`.

We can easily see that `(df)/dx >0` for `x <0` and `(df)/dx < 0` for `x > 0`, so the function will be strictly increasing from `x=-2` to `x=0` where it has a local maximum and strictly decreasing from `x=0` to `x=2`.

The value of the local maximum is `f(0) = 2` and it is clearly also the absolute maximum.

`(d^2f)/dx^2 = (-sqrt(4-x^2)-x^2/sqrt(4-x^2))/(x^2-4)`

`(d^2f)/dx^2 = (-4+x^2-x^2)/((x^2-4)sqrt(4-x^2))`

`(d^2f)/dx^2 = -4/((x^2-4)sqrt(4-x^2))`

The second derivative is always negative, so the function is concave down in its entire domain.

graph{sqrt(4-x^2) [-4.933, 4.932, -2.466, 2.467]}

If you let:

`y = sqrt(4-x^2)`

then:

`y^2 = 4-x^2`

`x^2+y^2 = 4`

so the graph is the part of the circle of center in the origin and radius `r = 2` above the `x` axis.